All the numbers known to humans except the imaginary numbers come under the set of the real number. We use R to denote the set of real numbers. We can have various subsets of the real number that denote different types of numbers. Various subsets of the Real number are,
Subsets of Real Numbers
Real Numbers can be divided into the following subsets:
Natural Numbers
Whole Numbers
Integers
Rational Numbers
Irrational Numbers
Now let’s learn about these numbers in detail further in the article. The image added below shows various types of numbers.
Natural Numbers
Natural numbers are the counting numbers i.e., numbers we use in our daily life for counting purposes, like 1, 2,3, 4, 5, and so on. As there can’t be -2 apples in the real world and also counting always starts with 1, not 0. So, only 1, 2, 3, 4, 5,… to infinity, are the only natural numbers, and these numbers are denoted by the symbol N.
Whole Numbers
Adding 0 to Natural Numbers makes the series a set of Whole Numbers. This collection of numbers {0, 1, 2, 3, 4, 5,…}, are known as whole numbers and denoted by the symbol W. The reason these numbers are called the whole number is that these numbers contain all the positive integers, which makes them complete, another name for complete is whole.
Integers
The collection of the whole number and negative natural numbers together is called integers; thus, integers include positive numbers, negative numbers, and one neutral number i.e., 0. {…, -3, -2, -1, 0, 1, 2, 3,…} are a representation of integers in collection form, other than these integers are denoted by the symbol Z.
Rational Numbers
Rational Numbers are numbers that can be expressed as the fraction p/q of two integers, a numerator p, and a non-zero denominator q such as 2/7. For example, 25 can be written as 25/1, so it’s a rational number. Some more examples of rational numbers are 22/7, 3/2, -11/13, -13/17, etc. As rational numbers cannot be listed in the collection form, the only representation of the rational numbers is Q.
Irrational Numbers
Irrational numbers are those which cannot be expressed in the form of p/q where p and q are both integers and q ≠ 0. In short, irrational numbers are real numbers that are not rational numbers. √2 is an irrational number as it can’t be written in p/q form i.e., in √2/1, √2 is not an integer. √3, √5, π, etc. are some more examples of irrational numbers.
List of Real Numbers
The list of real numbers is endless because it includes all kinds of numbers like whole, natural, integers, rational, and irrational numbers. Since it includes integers it has negative numbers too. So, there is no specific number from which the list of real numbers starts or ends. It goes to infinity towards both sides of the number line.
Symbol of Real Numbers
We use R to represent a set of Real Numbers and other types of numbers can be represented using the symbol discussed below,
N – Natural Numbers
W – Whole Numbers
Z – Integers
Q – Rational Numbers
Q’ – Irrational Numbers
Real Numbers Chart
Rational Numbers, Irrational Numbers, and all the subsets of rational numbers all come under real numbers, and the real number chart is used to express all the branches of real numbers. The real number chart is added below:
Properties of Real Numbers
There are different properties of Real numbers with respect to the operation of addition and multiplication, which are as follows:
A number line is a representation of Numbers with a fixed interval in between on a straight line. A Number line contains all the types of numbers like natural numbers, rational numbers, Integers, etc.
Representation of Rational Numbers on the Number Line
As shown in the following number line 0 is present in the middle of the number line. Positive integers are written on the right side of zero whereas negative integers are written on the left side of zero, and there are all possible values in between these integers.
Rational numbers are written between the numbers they lie. For example, 3/2 equals to 1.5, so is noted between 1 and 2. It shows that the number 3/2 lies somewhere between 1 and 2.
Similarly, the Number 13/4 = 3.25 lies between 3 and 4. So we noted it between 3 and 4. Number -50/9 = -5.555. . . , lies between -5 and -6. So we noted it between -5 and -6 on the number line.
Example: Represent the Following numbers on a number line:
23/5
6
-33/7
Solution:
The rational numbers,
23/5
6
-33/7
can easily be represented in a number line as,
Representation of Irrational Numbers on Number Line
Irrational Numbers can’t be represented on the number line as it is, we need clever tricks and geometry to represent irrational numbers on a number line. Learn more about the Representation of √3 on a number line.
Successive Magnification
The process of representation and visualization of real numbers on the number line through a magnifying glass is known as successive magnification.
Let’s take the example of 3.25.
In the first line, we see that 3.25 lies between 3 and 4. Now take it a step forward. Now we zoom in between 3 and 4 as represented in the second line. Here we found that 3.25 lies between 3.2 and 3.3. Now again, zoom in between 3.2 and 3.3 which is represented in the third line. Here we can see the required value of 3.25.
So, we represented 3.25 on the number line using successive magnification.
Decimal Expansion of Real Numbers
The decimal expansion of a real number is its representation in base equals to 10 (i.e., in the decimal system). In this system, each “decimal place” consists of a digit from 0 to 9. These digits are arranged such that each digit is multiplied by a power of 10, decreasing from left to right.
Decimal Expansion of Rational Numbers
Can we represent 13/4 in another form that can show its exact value on the number line?
The answer is yes. We can write it in decimals which gives its exact value. Let’s Expand 13/4
So 13/4 can also be written as 3.25.
Now Let’s take another example. Let’s expand 1/3
So 1/3 can also be written as 0.3333…… We can also write it as
Similarly, 1/7 can be written as 0.142857142857142857… or . This is known as the recurring decimals expansion.
Decimal Expansion of Irrational Numbers
Decimal Expansion of Irrational Numbers is non-terminating and non-repeating. We can find the decimal expansion such as √2, √3, √5, etc. using the long division method. The decimal Expansion of √2 is up to three digits after the decimal is calculated in the following illustration.
We can perform mathematical operations, such as Division, Multiplication, Addition, and Subtraction, on real numbers and the result still be a real number. Also, on the subset of real numbers, such as Rational numbers, we can perform all the operations between two rational numbers and the result will always be rational numbers, but on Irrational numbers, when we perform mathematical operations, it can be either rational or irrational.
For example, the product of 3/5 and 5/7 (two rationals) is 3/7 which is a rational number but the product of √2 and 3√2 (two irrationals) is 6 which is rational.
Sample Examples on Real Numbers
Example 1: Add √3 and √5
Solution:
(√3 + √5)
Now answer is an irrational number.
Example 2: Multiply √3 and √3.
Solution:
√3 × √3 = 3
Now answer is a rational number.
So we can say that result of mathematical operations on irrational numbers can be rational or irrational.
Now add a rational number with an irrational number.
Example 3: Add 2 and √5
Solution:
(2 + √5)
Now answer is an irrational number.
Example 4: Simplify the expression: (2 + √3)(5 + √3)
Solution:
(2 + √3)(5 + √3)
= 10 + 2√3 + 5√3 + 3
= 13 + 7√3
Which is an irrational number.
FAQs on Real Numbers
Q1: Define Real Numbers.
Answer:
Collection of all rational as well as irrational numbers is called real numbers.
Q2: What is the difference between a rational and an irrational number?
Answer:
The key difference between rational and irrational number is rational numbers can be represented in the form of p/q, where p and q are integers and q≠0. Whereas irrational numbers can’t be represented in the same form.
Q3: Are whole numbers and integers real numbers?
Answer:
Yes, all whole numbers and integers are real numbers as whole numbers and integers are the subsets of the rational numbers.
Q4: What is the symbol used to represent real numbers?
Answer:
The symbol used to represent real numbers is ℝ OR R.
Q5: What is a decimal representation of a real number?
Answer:
Decimal Representation of a real number can be either terminating, non-terminating but repeating, or non-terminating non-repeating as a real number contains all real numbers as well as irrational numbers.
Q6: How are real numbers ordered?
Answer:
Real Numbers are ordered based on their position on the number the real number on the left is always smaller than the real number on the right on the number line.
Q7: Are imaginary numbers real numbers?
Answer:
No, imaginary Numbers are not real numbers as they can’t be represented on the number line.
Euclid’s Division Lemma Definition
Euclid’s Division Lemma gives the relation between the various components of Division. It explains that for any two positive integers say a and b there exist two unique integers q and r such that a = bq + r. In this method, we call q the quotient of the division, and r is the remainder of the division.
This is nothing but another name for Euclid’s Division Lemma
Now let’s look at the statement of Euclid’s Division Lemma
Statement of Euclid’s Division Lemma
For any two positive integers a and b, there exists a unique set of integers q and r, such that:
a = b q + r
Where,
q is called the quotient, and
r is called the remainder, 0 ≤ r < b.
Let’s consider an example of Euclid’s Division Lemma for better understanding.
Here, the given numbers are, 39(=a) and 6(=b) we can write it in a = bq + r form as,
39 = 6×6 + 3
where, quotient(q) is 6 and remainder(r) is 3.
Now, the Euclid Division Lemma is, a = b × (q + r) can be written as,
Dividend = (Divisor × Quotient) + Remainder
Euclid’s Division Lemma has many Applications related to the Divisibility of Integers
It can be used to findthe HCF of two numbers.
Proof of Euclid’s Division Lemma
For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.”
To prove Euclid’s Division Lemma, we need to establish two main parts: the existence and uniqueness of the integers q and r.
Existence
We will prove the existence of integers q and r such that a = bq + r, where 0 ≤ r < b.
Consider the set S = {a – bk | k is an integer and a – bk ≥ 0}. S represents the set of all possible remainders when a is divided by b.
If S is empty, then there are no possible remainders greater than or equal to 0. However, this contradicts the fact that itself is an element of S (since a – b × 0 = a), and therefore, S cannot be empty.
Now, let’s assume that S contains a positive integer. In other words, there exists some k such that a – bk = r > 0. We can rewrite this equation as a = bq + r, where q = k and r is the remainder.
The only remaining case is when S contains negative integers only. Suppose a – bx < 0 for all integers x. This implies that a < bx for all x, which further implies a < b × 0. But this contradicts the fact that a is a positive integer. Hence, it is not possible for S to contain negative integers only.
Therefore, there must be a non-negative integer r in S. Hence, we have proven the existence part of Euclid’s Division Lemma.
Uniqueness
We will now prove the uniqueness of the integers q and r in Euclid’s Division Lemma.
Suppose there exist two pairs of integers (q₁, r₁) and (q₂, r₂) such that a = bq₁ + r₁ and a = bq₂ + r₂, where 0 ≤ r₁ < b and 0 ≤ r₂ < b.
Subtracting these two equations, we get:
bq₁ + r₁ – (bq₂ + r₂) = 0
Simplifying, we have:
b(q₁ – q₂) + (r₁ – r₂) = 0
Now, since b(q₁ – q₂) is divisible by b, the left-hand side of the equation is divisible by b. Therefore, (r₁ – r₂) must also be divisible by b.
But we know that 0 ≤ r₁ < b and 0 ≤ r₂ < b, which means the absolute difference between r₁ and r₂ is less than b. Therefore, the only way (r₁ – r₂) can be divisible by b is if r₁ = r₂.
Hence, we conclude that r₁ = r₂. Substituting this back into the original equations, we get:
a = bq₁ + r₁ = bq₂ + r₁
This implies bq₁ = bq₂, and since b is a positive integer, we can cancel it from both sides to get q₁ = q₂.
Therefore, we have shown that if (q₁, r₁) and (q₂, r₂) are two pairs of integers satisfying a = bq₁ + r₁ and a = bq₂ + r₂, then q₁ = q₂ and r₁ = r₂. This establishes the uniqueness part of Euclid’s Division Lemma.
Hence, we have proven both the existence and uniqueness of the integers q and r in Euclid’s Division Lemma.
What is Euclid’s Division Algorithm?
The process of finding the HCF of two numbers using Euclid’s Division Lemma is called “Euclid’s Division Algorithm”. In this process, Euclid’s Division Lemma is used in succession several times to find the HCF of the two numbers.
The image added below states Euclid’s Division Algorithm.
Euclid’s Division Algorithm is one of the most common division algorithms taught to students to divide any number. An algorithm is a set of instructions provided to perform a specific task it repeats certain steps to complete a task. While applying Euclid’s Division Lemma we came across the condition that, for two integers a and b (a<b), there exists two integers q and r such that,
a = bq + r
After studying the above relation we can say that any number k which is the common factor of a and b is the common factor of r also. We use this concept to find the HCF of two numbers.
Euclid’s Division Algorithm Example: Finding HCF
We can easily find the HCF of two numbers 135 and 275 by applying Euclid’s Division Lemma on these two numbers multiple times i.e. we can directly say that we can find the HCF of 135 and 275 by applying Euclid’s Division Algorithm. Follow the following steps to find the HCF.
Step 1: Use Euclid’s Division Lemma on 135 and 275
275 = 135(2) + 5
As we know that HCF is the highest factor of 135 and 275 and also any factor of 275 and 135 is also a factor of 5. So further applying Euclid’s Division Lemma on 135 and 5. (This process is repeated until the remainder r in Euclid’s Division Lemma becomes zero(0) and the remainder in the second last step is the required HCF of given two numbers.
Step 2: Repeating Euclid’s Division Lemma further
135 = 5(27) + 0
As now the remainder is zero we can say that the HCF of 135 and 275 is 5
Generalizing Euclid’s Division Algorithm
Now we can generalize Euclid’s Division Algorithm for finding the HCF of any number. Assume that we have to find the HCF of two random numbers a and b then their HCF is found by applying Euclid’s Division Lemma multiple times until the remainder obtained is zero (0). This can be understood as,
a = b(q1) + r1
b = r1(q2) + r2
r1 = r2(q3) + r3
⋮
rn−2 = rn−1(qn) + rn
rn−1 = rn(qn+1) + 0
Now the HCF of a and b is the remainder obtained in the second last step, i.e., the HCF (a, b) = rn
Shortcut Method for Euclid’s Division Algorithm
We also have a shortcut method of finding the HCF of two numbers this method uses the concept of division to find the HCF. We know that,
Dividend = Divisor × Quotient + Remainder
Now we divide the given numbers (say and b) accordingly the bigger number with the smaller number and we obtained a quotient and a remainder. Now this remainder becomes the divisor and the previous divisor becomes the dividend and further, the division process is carried out.
This process is repeated till the remainder of the division is zero. And the quotient when the remainder becomes zero is the HCF of the two numbers a and b.
The following illustration shows the calculation of HCF of 132 and 320 using the shortcut method.
Thus, the HCF of 132 and 320 is 4.
Solved Examples on Euclid’s Division Lemma
Example 1: Find the quotient and remainder when 73 is divided by 9 using Euclid’s Division Algorithm.
Solution:
Given:Dividend = 73, Divisor = 9
Using Euclid’s Division Lemma, Divide 73 by 9
⇒ 73 = 9 × 8 + 1
Therefore, when 73 is divided by 9, the quotient is 8 and the remainder is 1.
Example 2: Find the quotient and remainder when 315 is divided by 17 using Euclid’s Division Algorithm.
Solution:
Given:Dividend = 315, Divisor = 17
Using Euclid’s Division Lemma, Divide 315 by 17
⇒ 315 = 17 × 18 + 9
Thus, quotient is 18 and remainder is 9.
Example 3: Use Euclid’s Division Algorithm to Find the HCF of 867 and 255.
Solution:
Step 1: Since 867 > 255, we apply the Division Lemma on 867 and 255,
867 = 255 × 3 + 102
Step 2: Since the Remainder 102 ≠ 0, we apply the division lemma further on 255 and 102
255 = 102 × 2 + 51
This step is repeated until the remainder becomes zero.
Step 3: Since the Remainder 51 ≠ 0, we apply the division lemma further on 102 and 51
102 = 51 × 2 + 0
Now the remainder is zero, so our process stops. Since the divisor at this stage is 51, the HCF of 255 and 867 is 51.
Example 4: Use Euclid’s Division Algorithm to Find the HCF of 4052 and 12576.
Solution:
Step 1: Since 12576 > 4052, we apply the Division Lemma on 867 and 255,
12576 = 4052 × 3 + 420
Step 2: Since the Remainder 420 ≠ 0, we apply the division lemma to 4052 and 420,
4052 = 420 × 9 + 272
Step 3: Since the Remainder 272 ≠ 0, we apply the division lemma to 420 and 272
420 = 272 × 1 + 148
This step is repeated until the remainder becomes zero.
420 = 272 × 1 + 148
Since the Remainder 148 ≠ 0, we apply the division lemma to 272 and 148.
272 = 148 × 1 + 124
Since the Remainder 124 ≠ 0, we apply the division lemma to 148 and 124.
148 = 124 × 1 + 24
Since the Remainder 24 ≠ 0, we apply the division lemma to 124 and 24.
124 = 24 × 5 + 4
Since the Remainder 4 ≠ 0, we apply the division lemma to 24 and 4.
24 = 4 × 6 + 0
Now the remainder becomes zero, so our process stops. Since the divisor at this stage is 4, the HCF of 4052 and 12576 is 4.
FAQs on Euclid’s Division Algorithm
Q1: What is Euclid’s Division Lemma?
Answer:
Euclid’s Division Lemma is the basic theorem of mathematics that says for any two positive integers a and b, we have two integers q and r such that,
a = bq + r
Q2: What is Euclid’s Division Algorithm?
Answer:
Euclid’s Division Algorithm is an algorithm which is used to find the HCF of two numbers by using Euclid’s Division Lemma in succession till the remainder(r) is made zero.
Q3: What is the difference between Euclid’s Division Lemma and Euclid’s Division Algorithm?
Answer:
Euclid’s Division Lemma is the basic theorem of Euclid’s Division whereas Euclid’s Division Algorithm is an Algorithm which uses Euclid’s Division Lemma to find the HCF of two numbers.
Q4: What is the use of Euclid’s Division Algorithm?
Answer:
Euclid’s Division Algorithm is used to find the HCF of two or more numbers.
Q5: What is the formula of the Division Algorithm?
Answer:
The formula of the Division Algorithm is,
Dividend = Divisor × Quotient + Remainder
Q6: How does Euclid Algorithm calculate HCF?
Answer:
The HCF of two numbers a and b can be calculated using Euclid Algorithm by following the steps discussed below,
Step 1: Use Euclid’s Division Lemma, a and b. So, we find q and r such that a = bq + r, 0 ≤ r < b.
Step 2: If r = 0, b is the HCF of a and b. If r ≠ 0, repeat step 1 till the remainder becomes zero.
Step 3: In the case when the remainder is zero, the divisor b is the HCF of a and b.
Fundamental Theorem of Arithmetic
Let’s take some set of prime numbers, for example — {3, 2, 7}. How many numbers do you think we can make from their multiplication? 3 × 2 = 6, 3 × 3 × 2 = 18, 7 × 2 = 14 and so on. So, we can say, infinitely many numbers can be made from these prime numbers. But does that prove that we can generate every possible number?
Yes, there are infinitely many possible prime numbers and from their multiplication, we can generate infinite numbers and that is the crux of the Fundamental Theorem of Arithmetic. To further develop this concept let’s look at the factorization of a number. Suppose we are given a number x = 36.
The figure above represents the factorization tree of the number. 36 = 2 × 2 × 3 × 3. It is a product of prime numbers. If we keep on trying different numbers, we see that all the numbers can be represented as product primes. In a more formal way,
Theorem:
Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
This is called Fundamental Theorem of Arithmetic.
This theorem says that every composite number can be rewritten as the product of prime numbers in a “unique” way, except for the order in which primes occur.
Proof of Fundamental Theorem of Arithmetic
Step 1: The existence of prime factors, we will prove it b y induction
Firstly, consider n>1
Therefore Initially, n=2. Since n=2 and 2 is a prime number, the result is true.
Consider n>2 (Induction hypothesis: Let result be true for all positive numbers less than n )
No, we will prove that the result is also true for n.
If n is prime, then n is a product of primes is trivially true.
If n is not prime i.e n is a composite number, then
n = ab, a, b < n
By induction method, the result is true for a and b (because a<n and b<n). Therefore, by the induction hypothesis, a must be the product of prime numbers and b is a product of prime numbers. Therefore, n = ab is a product of prime numbers. Thus, it is proved by induction.
Step 2: Uniqueness (of factors up to order)
Let n = p1 p2 p3….pk (where p1 p2 p3 …pk are primes)
Let if possible, there be two representations of n as a product of primes
i.e let if possible n=p1 p2 p3….pk = q1 q2q3 …qr where pi’s and qj’s are prime numbers
So by the same argument, we will get p2=q2 and so on.
Thus, n can be expressed as a product of primes uniquely (except for the order)
Hence proved.
Question 1: Factorize the number “4072” and represent it in tree form.
Answer:
Question 2:Factorize the number “324” and represent it in tree form.
Answer:
Question 3:Factorize the number “16048” and represent it in tree-form.
Answer:
LCM and HCF using Fundamental Theorem of Arithmetic
HCF known as the highest common factor is the greatest number that divides each of the two numbers given.
LCM is the lowest common multiple that is the product of all the common prime factors but with their highest degrees/powers.
For example:
Question 1: Find the LCM and HCF of 24 and 36.
Solution:
The Prime factors of 24 = 2× 2×2×3
The prime factors of 36 = 2×2×3×3
HCF = 2×2×2×3, 2×2×3×3 = 2×2×3 = 12
LCM =
2×2×2×3×3 = 72
LCM and HCF can be found with the help of prime factorization too, lets’s look at some examples.
Question 2: Find the LCM and HCF of numbers 6 and 20.
Answer:
Prime Factorization of 6 can be represented in the following way,
Prime Factorization of 20 can be represented in the following way,
So, now we have prime factorization of both the numbers,
6 = 2 × 3
20 = 2 × 2 × 5
We know that
HCF = Product of the smallest power of each common prime factor in the numbers.
LCM = Product of the greatest power of each prime factor, involved in the numbers.
So, HCF(6,20) = 21
LCM(6,20) = 22 × 31 × 5
Question 3:Find the LCM and HCF of numbers 24 and 36.
Answer:
Prime Factorization of 24:
Prime Factorization of 36:
24 = 23 × 3 and 36 = 22 × 32
Based on the previous definitions,
HCF(24, 36) = 12
LCM(24, 36) = 72
Fact: In the above examples, notice that for any two numbers “a” and “b”. HCF × LCM = a × b.
Question 4: Suppose that for two numbers “a” and “b”. HCF is given which is 120 and the product of the two numbers is given as 3600. Find the LCM of the two numbers.
Answer:
Given two numbers “a” and “b”.
LCM(a, b) is unknown while HCF(a, b) = 120 and a × b = 3600.
From the property studied above,
HCF(a, b) × LCM(a, b) = a × b
Plugging in the given values.
120 × LCM(a, b) = 3600
LCM(a, b) = 30
How to Find HCF and LCM?
There are various methods to find HCF and LCM. Following are some of the most famous methods used to calculate the Highest Common factor and Least Common Multiple.
Division method
Factorization method
Prime factorization method
HCF by Division Method
The easiest way to understand how to find HCF by Division Method is by going back to simple division. Following are the steps for better understanding,
Step 1: Take the smaller number as the divisor and the larger number as a dividend.
Step 2: Perform division. If you get the remainder as 0, then the divisor is the HCF of the given numbers.
Step 3: If you get a remainder other than 0 then take the remainder as the new divisor and the previous divisor as the new dividend.
Step 4: Perform steps 2 and step 3 until you get the remainder as 0.
Example: Find out the HCF of 36 and 48.
Solution:
Using the division method for HCF
Hence, HCF = 12
LCM by Division Method
In order to find LCM formula by Division method, we divide the Following steps can be followed in order to find Least Common Division by Division Method:
Step 1: Check whether the given numbers are divisible by 2 or not.
Step 2: If the number is divisible by 2 then divide and again check for the same. If the numbers are not divisible by 2 then check 3, and so on.
Step 3: Perform step 2 until you get 1 in the end.
Example: Find out the LCM of 36 and 48.
Solution:
Using the division method for LCM
Hence, LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
HCF by Factorization Method
Finding HCF by Factorization method requires the given steps to be followed:
Step 1: Write all the divisors of the given number.
Step 2: Check for common divisors among them and find the greatest common divisor. This greatest common divisor will be the HCF of the given numbers.
Example: Find out the HCF of 6 and 18.
Solution:
Divisors of 6 = 1, 2, 3, 6
Divisors of 18 = 1, 2, 3, 6, 9, 18
HCF = greatest common divisor
HCF = 6
LCM by Factorization Method
The factorization method for LCM requires minimum steps; below are the steps mentioned:
Step 1: Write the multiples of the given numbers until you reach the first common multiple.
Step 2: First common multiple of the given numbers will be the LCM.
Example:Find out the LCM of 6 and 18.
Solution:
Multiple of 6 = 6, 12, 18, 24, 30, …
Multiple of 18 = 18, 36, 54, …
LCM = first common multiple (least common multiple)
LCM = 18
HCF by Prime Factorization
Finding HCF by Prime Factorization can be done by following the given steps:
Step 1: Find out the prime factors of the given number.
Step 2: Check the occurrence of a particular factor. Find out the common factors and choose them in HCF.
Step 3: Multiply the occurrence of common factors. And this will be the HCF Of the given numbers.
Example: Find out the HCF of 18 and 90.
Prime factors of 18 = 2 × 3 × 3
Prime factors of 90 = 2 × 3 × 3 × 5
Now, HCF = 2 × 3 × 3 = 18
LCM by Prime Factorization
Finding LCM by Prime Factorization is done by following the given steps:
Step 1: Find out the prime factors of the given number.
Step 2: Check the occurrence of a particular factor. If a particular factor has occurred multiple times in the given number, then choose the maximum occurrence of the factor in LCM. It can also be found out by checking the powers of the factors. The factor having greater power will be chosen between the numbers.
Step 3: Multiply all the maximum occurrences of a particular factor. And this will be the LCM Of the given numbers.
Example: Find out the LCM of 18 and 90.
Solution:
Prime factors of 18 = 2 × 3 × 3
Prime factors of 90 = 2 × 3 × 3 × 5
Now, LCM = 2 × 3 × 3 × 5 = 90
Alternate method:
Prime factors of 18 = 2 × 3 × 3
Prime factors of 18 = 21 × 32
Prime factors of 90 = 2 × 3 × 3 × 5
Prime factors of 90 = 21 × 32 × 51
Chosen factors for LCM = 21 × 32 × 51
Therefore, LCM = 2 × 9 × 5 = 90.
HCF and LCM Formula
In order to find the HCF and LCM formula, let’s assume that the numbers given are a and b. The relationship between HCF and LCM states that the product of a and b is equal to the product of HCF and LCM.
(LCM of two numbers) × (HCF of two numbers) = Product of two numbers
Mathematically this can be written as:
LCM(a, b) × HCF(a, b) = a × b
Difference between HCF and LCM
HCF
LCM
It is the Highest Common Factor.
It is the Least Common Multiple.
The greatest of all the common factors among the given numbers is HCF.
The smallest of all the common multiples among the given numbers is LCM.
The HCF of given numbers will never be greater than any of the numbers.
The LCM of the given numbers will always be greater than the numbers given.
Example 1: Find out the LCM and HCF of 18, 30, and 90 by prime factorization.
Solution:
Prime factors of 18 = 2 × 3 × 3
Prime factors of 30 = 2 × 3 × 5
Prime factors of 90 = 2 × 3 × 3 × 5
LCM: 2 × 3 × 3 × 5 = 90
HCF: 2 × 3 = 6
Example 2: Find out the LCM and HCF of 318 and 504.
Solution:
Prime factors of 16 = 2 × 2 × 2 × 2
Prime factors of 30 = 2 × 3 × 5
LCM: 2 × 2 × 2 × 2 × 3 × 5
HCF: 2
Example 3: Find out the HCF of 24 and 36.
Solution:
Let’s find out the HCF of 24 and 36 by division method,
Therefore,
HCF = 2 × 2 × 3 = 12
Example 3: Find out the LCM of 24 and 36.
Solution:
Let’s find out the LCM pf 24 and 36 by division method,
Therefore,
LCM = 2 × 2 × 3 × 2 × 3 = 72
Example 4:Find out the LCM and HCF of 15 and 70. Also, verify the relationship between LCM, HCF, And given numbers.
Solution:
Prime factors of 15 = 3 × 5
Prime factors of 70 = 2 × 5 × 7
LCM: 2 × 3 × 5 × 7
HCF: 5
Verifying the relationship:
LCM × HCF = 2 × 3 × 5 × 5 × 7 = 1050
Product of two numbers = 15 × 70 = 1050
From above you can see that
LCM (15, 70) × HCF(15, 70) = Product of two numbers
Hence Verified.
FAQs on HCF and LCM
Q1: State the full form of HCF and LCM.
Answer:
The full form of HCF is the Highest Common Factor and the full form of LCM is the Lowest Common Multiple. For example, the HCF of 4 and 12 is 4, the LCM of 3 and 10 is 30.
Q2: What is the relationship between HCF and LCM?
Answer:
The relationship between HCF and LCM states that the product of a and b is equal to the product of HCF and LCM.
(LCM of two numbers) × (HCF of two numbers) = Product of two numbers
This is written as:
LCM(a, b) × HCF(a, b) = a × b
Q3: What is GCF? Find GCF of 4 and 12.
Answer:
GCF is the Greatest Common Factor which is nothing but another name for HCF.
GCF of 4 and 12 will be 4 as 4 is the greatest common factor between 4 and 12.
Q4: What are the different methods to find HCF and LCM in math?
Answer:
There are various methods to find HCF and LCM. Following are some of the most famous methods used to calculate the Highest Common factor and Least Common Multiple.
Division method
Factorization method
Prime factorization method
Q5: What is the use of HCF and LCM?
Answer:
HCF is very useful in maths and in real life as well. When numbers are required to be divided into smaller sections, or when certain things are required to be into smaller parts or in groups, HCF is used. LCM can be used in places where certain situations will require to occur multiple times.
Irrational Number Definition
Irrational Numbers are numbers which can not be expressed as the ratio of two integers. They are a subset of Real Numbers and can be expressed on the number line. And, the decimal expansion of an irrational number is neither terminating nor repeating.
We can define irrational numbers as real numbers that cannot be expressed as p/q where p and q are integers and q ≠ 0.
We know that irrational numbers are real numbers and they cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0. For example, √ 5 and √ 3, etc. are irrational numbers. On the other hand, the numbers which can be represented in the form of p/q, such that, p and q are integers and q ≠ 0, are rational numbers.
Examples of Irrational Numbers
√2, √3, π, e are some examples of irrational numbers.
√2 = 1.41421356237309504880…
Pi “π”= The value of π is 3.14159265358979323846264338327950… It is a really famous irrational number. People have calculated its value up to quadrillion decimals but still haven’t found any pattern yet.
Euler’s number“e” = Euler number is also very popular in mathematics. In this case, also, people tried calculating it up to a lot of decimals but still, no pattern was found. the value of e = 2.7182818284590452353602874713527 (and more …).
Golden ratio “ϕ” = This is an irrational number and its application is found in many fields like computer science, design, art, and architecture.
Irrational Number Symbol
We represent the Irrational number with the symbol Q’ as Q represents the group of rational numbers so Q complement (Q’) is used to represent irrational numbers. Also,
Q U Q’ = R
where R is the set of real numbers.
How to know a number is Irrational?
We know that rational numbers are expressed as, p/q, where p and q are integers and q ≠ 0. But we can not express the irrational number in a similar way. Irrational numbers are non-terminating and non-recurring decimal numbers. So if in a number the decimal value is never ending and never repeating then it is an irrational number.
Some examples of irrational numbers are,
1.112123123412345…
-13.3221113333222221111111…, etc.
Are Irrational Numbers Real Numbers?
Irrational numbers come under real numbers, i.e. all irrational numbers are real. But irrational numbers are different from rational numbers as they can’t be written in the form of fractions. Although, irrational numbers can be expressed in the form of non-terminating and non-recurring fractions. For example, √2, √3, and π are all irrational numbers and can’t be written as fractions. The image below explains the relationship between Irrational numbers and Real Numbers.
Properties of Irrational Numbers
Various properties of irrational numbers are discussed below:
Sum of two rational numbers may be rational or may be irrational.
Sum of a rational number and an irrational number is an irrational number.
Product of an irrational number with a non-zero rational number is an irrational number.
Product of two irrational numbers may be rational or may be irrational.
LCM of two irrational numbers may or may not exist.
Set of irrational numbers is not closed under the multiplication process, but a set of rational numbers is closed.
Product of Two Irrational Numbers
Product of two rational numbers may be either rational or irrational. For example,
π × π = π2 is irrational.
√2 × √2 = 2 is rational.
So Product of two Irrational Numbers can result in a Rational or Irrational Number accordingly
Sum of Two Irrational Numbers
The Sum of two irrational numbers is sometimes rational sometimes irrational.
3√2 + 4√3 is irrational.
(3√2 + 6) + (- 3√2) = 6, is rational.
Product of Irrational Number and Non-zero Rational Number
The product of any irrational number with any non-zero rational number is an irrational number.
For example, 3×√2 is an irrational number as it can not be represented as p/q
Fun Facts
Apparently, Hippasus (one of Pythagoras’ students) discovered irrational numbers when trying to write the square root of 2 as a fraction (using geometry, it is thought). Instead, he proved the square root of 2 could not be written as a fraction, so it is irrational. But followers of Pythagoras could not accept the existence of irrational numbers, and it is said that Hippasus was drowned at sea as a punishment from the Gods!
Is Pi an Irrational Number?
Yes, Pi (π) is an irrational number because it is neither terminating decimals nor repeating decimals. We will learn more about Pi (π) as,
Let’s take a circle, measure its circumference, and divide it by its diameter. It will always be a constant if measured accurately.
This constant ratio is denoted by the Greek symbol π (read as pi). That is,
Circumference/diameter = π
It is an important universal constant, it occurs in lots of places in our universe and in our daily lives. It was not created by humans, it was discovered.
What is the value of Pi
π = 3.14159265358979323846264338327950…
It is not infinite, it is an irrational number.
Note: We often take 22/7 as the value of Pi, but it is an approximation.
Now one might think, how is Pi irrational? One can measure the circumference, One can measure the diameter, and then take their ratio. So it must be rational as we define π as,
π = Circumference/Diameter
And if we define any number as that ratio of two numbers then it is a rational number. But while defining the ratio of π, we take approximation and the value of π is never exactly equal to the ratio of Circumference and Diameter.
Yes, √2 is an irrational number because it is neither terminating decimals nor repeating decimals. But we can graphically represent the value of √2 as the diagonal of the square with sides one unit as shown in the image below.
There are various irrational numbers that are widely used in mathematics. Some of the most commonly used irrational numbers are discussed in the table below,
Irrational Number
Symbol
Approximate Value
Pi
π
3.14159265358979…
Euler’s Number
e
2.71828182845904…
Golden Ratio
φ
1.61803398874989….
Irrational Number Theorem
The irrational number theorem is,
Theorem: If p is a prime number and p divides a2, then p also divides a.
Proof:
Using the Fundamental Theorem of Arithmetic
We can say that,
a = p1 × p2 × p3……….. × pn …..(i)
where, p1, p2, p3, ……, pn are prime factors of a.
Squaring both sides of equation (ii)
a2 = (p1 × p2 × p3……….. × pn)2
a2 = (p1)2 × (p2)2 × (p3)2…….. × (pn)2
According to the Fundamental Theorem of Arithmetic. Every natural number has a unique prime factor.
Prime factors of a2 are p1, p2, p3……….., pn. Also, the prime factors of a are, p1, p2, p3……….., pn
Thus, if a2 is divisible by p, then p also divides a.
How to Find Irrational Numbers?
An irrational Number between two prime numbers say a and b is given by √ab.
Irrational numbers between any two numbers are also found using the concept of perfect squares. We know that,
√(1) = 1
√(4) = 2
As the perfect squares of the number between them do not exist. So the numbers between them are irrational numbers, i.e. √(2), and √(3) are irrational numbers. Similarly the numbers
√(4) = 2
√(9) = 3
So the numbers between them, √(5), √(6), √(7), and √(8) are irrational numbers.
Also, the cube root of the non-perfect cube is an irrational number.
So we can say that the root of prime numbers is irrational numbers. √P is an irrational number where P is the prime number.
This can be proved using the contradiction method.is
Statement: Square root of the prime number is an irrational number.
Proof:
Let us assume that, √u is a rational number.
By the definition of rational numbers
√u =p/q …….(i)
Where p and q have no common factor other than 1 and q ≠ 0
Squaring both sides of equation (1), we have
u = p2/q2
p2 = uq 2 ………. (ii)
Now we can say that if u is a prime factor of p2, then u is a prime factor of p.
Thus,
p = u × c, where c is any integer.
from eq (ii)
(uc)2 = uq2
q2 = uc2…(iii)
From eq(iii) we can say that if u is a prime factor of q2, then u is a prime factor of q.
But initially we have assumes that p and q have no common factors. But from the above proof, we say that p and q have a common factor u, which implies that our initial assumption is wrong. That is √u is not a rational number. Thus, √u is an irrational number.
Differences Between Rational and Irrational Numbers
The difference between rational and irrational numbers is discussed in the table below.
Rational numbers
Irrational numbers
We express rational numbers as the ratio of integers i.e. rational numbers are expressed as p/q where q ≠ 0
We can not express the irrational number as a ratio of two integers.
Decimal expansion of rational numbers is recurring and terminating decimals.
Decimal expansion of irrational numbers are non-recurring and non-terminating decimals.
Examples of Rational Numbers: 2.345, 12.6565.., 1.75
Example 1: Find Rational Numbers or Irrational Numbers among the following.
2, 3, √3, √2, 1.33333…, 1.1121231234…
Solution:
Rational Numbers: 2, 3, 1.3333…. are rational numbers
Irrational Numbers: √3, √2, 1.1121231234… are irrational numbers
Example 2: Find the sum of the following irrational numbers.
a) √2, √2 b) √2, √3
Solution:
a) √2 + √2 = 2√ 2 (they are added as two like variables)
b) √2 + √3 = √2 + √3 (they can’t be added as unlike variables)
Example 3: Find the product of the following rational numbers.
a) √2, √2 b) √2, √3
Solution:
a) √2 × √2 = 2
b) √2 × √3 = √6
FAQs on Irrational Numbers
Q1: What is the Irrational Number definition?
Answer:
Irrational numbers are the numbers that are not defined as p/q where p and q are integers with no common factors and q does not equal to zero. They are non-terminating and non-recurring decimal numbers whose exact value can never be found.
Q2: What are Examples of Irrational numbers?
Answer:
The examples of rational number are,
√2 = 1.414…
√3 = 1.732…
Pi (π) = 3.1415…
Q3: Are Integers Irrational Numbers?
Answer:
No, Integers are not irrational numbers but are rational numbers as they can easily be represented in, p/q form. For example integers like, -1 and 2 are represented in rational number form as,
-1 = -1/1
2 = 2/1
Q4: How can you Identify an Irrational Number?
Answer:
Irrational number are the numbers which are non-recurring and non-terminating decimals and we can easily identify them using this property. For example,
1.123123123…. is a rational number as it is a recurring decimal.
1.123123 is a rational number as it is a terminating decimal.
1.1121231234…. is an irrational number as it is non-recurring and non-terminating decimal.
Q5: Why Pi is an Irrational Number?
Answer:
Pi(π) is an irrational number because it can not be expressed in the form of p/q and it is a non-recurring and non-terminating decimal and its value is,
π = 3.14156162…
Q6: How many Irrational Numbers are between 2 and 3?
Answer:
There are infinite irrational number between any two real numbers and thus, there are infinite irrational number between 2 and 3.
Rational Numbers
A number of the form p/q, where p and q are integers and q ≠ 0 are called rational numbers.
Examples:
1) All natural numbers are rational,
1, 2, 3, 4, 5…….. all are rational numbers.
2) Whole numbers are rational.
0,1, 2, 3, 4, 5, 6,,,,,, all are rational.
3) All integers are rational numbers.
-4.-3,-2,-1, 0, 1, 2, 3, 4, 5,,,,,,,, all are rational numbers.
Irrational Numbers
The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational numbers.
1) If m is a positive integer which is not a perfect square, then √m is irrational.
√2 ,√3, √5, √6, √7, √8, √10,….. etc., all are irrational.
2) If m is a positive integer which is not a perfect cube , then 3√m is irrational.
3√2, 3√3, 3√4,….. etc., all are irrational.
3) Every Non Repeating and Non Terminating Decimals are Irrational Numbers.
0.1010010001…… is an non-terminating and non repeating decimal. So it is irrational number.
0.232232223…….. is irrational.
0.13113111311113…… is irrational.
Nature of the Decimal Expansions of Rational Numbers
Theorem 1: Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then, x is a terminating decimal only when q is of the form (2m x 5n) for some non-negative integers m and n.
Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then, x is a nonterminating repeating decimal, if q ≠ (2m x 5n).
Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q = 2m x 5n then x has a decimal expansion which terminates.
Proof 1: √2 is irrational
Let √2 be a rational number and let its simplest form is p/q.
Then, p and q are integers having no common factor other than 1, and q ≠ 0.
Now √2 = p/q
⇒ 2 = p2/q2 (on squaring both sides)
⇒ 2q2 = p2 ……..(i)
⇒ 2 divides p2 (2 divides 2q2 )
⇒ 2 divides p (2 is prime and divides p2 ⇒ 2 divides p)
Let p = 2r for some integer r.
putting p = 2r in eqn (i)
2q2 = 4r2
⇒ q2= 2r2
⇒ 2 divides q2 (2 divides 2r2 )
⇒ 2 divides p (2 is prime and divides q2 ⇒ 2 divides q)
Thus 2 is common factor of p and q. But, this contradict the fact that a and b have common factor other than 1. The contradiction arises by assuming that √2 is rational. So, √2 is irrational.
Proof 2: Square roots of prime numbers are irrational
Let p be a prime number and if possible, let √p be rational.
Let its simplest form be √p=m/n, where m and n are integers having n no common factor other than 1, and
n ≠0.
Then, √p = m/n
⇒ p = m2/n2 [on squaring both sides]
⇒ pn2 = m2 ……..(i)
⇒ p divides m2 (p divides pn2)
⇒ p divides m (p is prime and p divides m2 ⇒ p divides m)
Let m = pq for some integer q.
Putting m = pq in eqn (i), we get:
pn2 = p2q2
⇒ n2 = pq2
⇒ p divides n2 [ p divides pq2]
⇒ p divides n [p is prime and p divides n2 = p divides n].
Thus, p is a common factor of m and n. But, this contradicts the fact that m and n have no common factor other than 1. The contradiction arises by assuming that /p is rational. Hence, p is irrational.
Proof 3: 2 + √3 is irrational.
If possible, let (2 + √3) be rational. Then, (2 + √3) is rational, 2 is rational
⇒ {( 2 + √3) – 2} is rational [difference of rationales is rational]
⇒ √3 is rational. This contradicts the fact that √3 is irrational.
The contradiction arises by assuming that (2 + √3) is irrational.
Hence, (2 + √3) is irrational.
Proof 4: √2 + √3 is irrational.
Let us suppose that (√2 + √3 ) is rational.
Let (√2 + √3) = a, where a is rational.
Then, √2 = (a – √3 ) ………….(i)
On squaring both sides of (i), we get:
2 = a2 + 3 – 2a√3 ⇒ 2a√3 = a2 + 1
Hence, √3 = (a² +1)/2a
This is impossible, as the right hand side is rational, while √3 is irrational. This is a contradiction.
Since the contradiction arises by assuming that (√2 + √3) is rational, hence (√2 + √3) is irrational.
Identifying Terminating Decimals
To Check Whether a Given Rational Number is a Terminating or Repeating Decimal Let x be a rational number whose simplest form is p/q, where p and q are integers and q ≠ 0. Then,
(i) x is a terminating decimal only when q is of the form (2m x 5n) for some non-negative integers m and n.
(ii) x is a nonterminating repeating decimal if q ≠ (2m x 5n).
Examples
(i) 33/50
Now, 50 = (2×52) and 2 and 5 is not a factor of 33.
So, 33/ 50 is in its simplest form.
Also, 50 = (2×52) = (2m × 5n) where m = 1 and n = 2.
53/343 is a terminating decimal.
33/50 = 0.66.
(ii) 41/1000
Now, 1000 = (23x53) = and 2 and 5 is not a factor of 41.
So, 41/ 1000 is in its simplest form.
Also, 1000 = (23x23) = (2m × 5n) where m = 3 and n = 3.
4 /1000 is a terminating decimal.
41/1000 = 0.041
(iii) 53/343
Now, 343 = 73 and 7 is not a factor of 53.
So, 53/ 343 is in its simplest form.
Also, 343 =73 ≠ (2m × 5n) .
53 /343 is a non-terminating repeating decimal.
How to Identify Rational Numbers?
All the rational numbers follow the following rules, thus using the help of these rules we can identify the rational numbers
Rational numbers are represented in the form of p/q, where q≠0.
Ratio p/q can be further simplified in simple form or decimal expansion.
Non-terminating decimals with repeating decimal values are also considered rational numbers as they can be represented in the form of p/q.
Example: Which of the following numbers are rational numbers?
a) -1.75 b) 2/3 c) √5 d) π
Solution:
a) -1.75 is a rational number as it it has a terminating decimal expansion.
b) 2/3 is also a rational number as it can be expressed in the form of a ratio of two integers.
c) √5 is an irrational number because it has a decimal expansion with infinitly many digits without any repetation.
d) π is also an irrational number as it has a decimal expansion with infinitly many digits without any repetation.
Thus, only (a) and (b) are the rational numbers out of all the given numbers.
Types of Rational Numbers
Various types of numbers can be represented as rational numbers some of which are discussed below:
Integers: All the integers, i.e. negative integers and positive integers both come under rational numbers. For example, -1, -11, -4, 6. 8. 10. etc all are rational numbers.
As natural numbers, whole numbers, and others come under rational numbers they are all rational numbers.
Fraction Number: A rational number is a ratio of two integers that can be written in the form of p/q where q is not equal to zero. Hence, any fraction with a non-zero denominator is a rational number.
Example: -2 / 5 is a rational number where -2 is an integer being divided by a non-zero integer 5
Decimal Number: A rational number can be also written in the decimal form if the decimal value is definite or has repeating digits after the decimal point.
Example: 0.3 is a rational number. The value 0.3 can be further expressed in the form of a ratio or fraction as p/q
0.3 = 3/10
Also, 1.333333… can be represented as 4/3 hence, 1.33333… is a rational number.
The image added below shows types of Rational Numbers.
Difference Between Fractions and Rational Numbers
Fractions are the real numbers represented in the form of a/b where both a and b are whole numbers whereas rational numbers are the real numbers represented in the form of a/b where both a and b are integers. However, in both cases, the denominator should not be equal to 0. Thus, we can say that all fractions are rational numbers but rational numbers are not fractions.
Operations on Rational Numbers
There are four most common operations for Rational Numbers, which are as follows:
Addition
Subtraction
Multiplication
Division
Addition
The addition of two rational numbers can be done using the following step-by-step method where the addition of 3/4 and 1/6 is explained as an example.
Step 1: Find the common denominator (LCD) for both the rational number. i.e.,
The common denominator for 4 and 6 is 12.
Step 2: Convert both the rational number to equivalent fractions using the common denominator. i.e.,
3/4 = (3 x 3)/(4 x 3) = 9/12
and 1/6 = (1 x 2)/(6 x 2) = 2/12
Step 3: Add the numerators of the equivalent fractions obtained in step 2. i.e.,
9/12 + 2/12 = (9 + 2)/12 = 11/12
Step 4: Simplify the resulting fraction if possible. i.e.,
11/12 is already in its simplest form.
Thus, Addition of 3/4 and 1/6 is 11/12 .
Subtraction
Subtraction of two Rational Numbers can be done using the following step-by-step method where subtraction of 1/3 and 2/5 is explained.
Step 1: Find the common denominator (LCD) for both the rational number. i.e.,
The common denominator for 3 and 5 is 15.
Step 2: Convert both the rational numbers to equivalent fractions with the common denominator. i.e.,
1/3 = (1 x 5)/(3 x 5) = 5/15
and 2/5 = (2 x 3)/(5 x 3) = 6/15
Step 3: Subtract the numerators of the equivalent fractions obtained in step 2. i.e.,
5/15 – 6/15 = (5 – 6)/15 = -1/15
Step 4: Simplify the resulting fraction if possible. i.e.,
-1/15 is already in its simplest form.
Therefore, 1/3 – 2/5 = -1/15.
Multiplication
Multiplication of two rational numbers can be achieved by simply multiplying the numerator and denominator of the given Rational Numbers. Step by step method with an example of multiplication of -11/3 and 4/5 is as follows:
Step 1: Write both rational number in with multiplication sign(×) in between. i.e.,
-11/3 × 4/5
Step 2: Multiply the numerator and denominator individually. i.e.,
(-11 × 4)/(3 × 5)
Step 3: We get the result of the multiplication. i.e.,
-44/15
Division
Division of two Rational numbers can be achieved in the following steps(where the division of 3/5 and 4/7 is explained):
Step 1: Write both rational number in with division sign in between. i.e.,
3/5 ÷ 4/7
Step 2: Change “÷” with “×” and take raciprocal of the second rational number. i.e.,
3/5 × 7/4
Step 3: Multiply the numerator and denominator of the resulting fractions. i.e.,
(3 × 7)/(5 × 4)
Step 4: We get the result of the division. i.e.,
21/20
The image added below shows all the properties of the rational numbers.
Rational Numbers Properties
Rational Numbers show several properties under the different operations (two of such common operations are addition and multiplication), which are as follows:
Closure Property
Commutative Property
Associative Property
Identity Property
Inverse Property
Distributive Property
Closure Property
Closure Property for Addition: Rational numbers are closed under addition, i.e., for any two rational numbers a and b, the sum a + b is also a rational number.
Closure Property for Multiplication: Rational numbers are closed under multiplication, i.e., for any two rational numbers a and b, their product ab is also a rational number
Example:
For a = 3 / 4 and b = (-1) / 2
Now, a + b = 3 / 4 + (-1) / 2
⇒ a + b = (3- 2)/ 4
⇒ a + b = 1/4, is Rational Number.
Also, a × b = 3/4 × (-1)/2 = -3/8, which is also Rational number.
Commutative Property
Commutative Property for Addition: Rational numbers hold commutative property under addition operation, i.e., for any two rational numbers a and b, a + b = b + a.
Commutative Property for Multiplication: Rational numbers hold commutative property under multiplication operation as well, i.e., for any two rational numbers a and b, ab = ba.
Example:
For a = (-7) / 8 and b = 3 / 5 Now, a + b = -7/8 + 3/5
⇒ a + b = (-7 x 5 + 3 x 8)/40 = (-35 + 24) / 40
⇒ a + b = (-11) / 40
And, b + a = 3/5 + (-7)/8
⇒ b + a = (3 x 8 + (-7) x 5)/ 40 = (24 – 35)/40
⇒ b + a = -11/40 = a + b
Now, ab = (-7)/8 x 3/5 = (-7 x 3)/(8 x 5)
⇒ ab = -21/40
And, ba = 3/5 x (-7)/8 = (3 x 7 )/(5 x 8)
⇒ ba =(-21)/40 = ab
Associative Property
Associative Property for Addition: Rational Numbers are associative under addition operation, i.e., for any three Rational Numbers a, b, and c, a + (b + c) = (a + b) + c
Associative Property for Multiplication: Rational Numbers are associative under multiplication operation as well, i.e., for any three Rational numbers a, b, and c, a(bc) = (ab)c
Example:
For three Rational numbers a,b,c where a = -1/2, b = 3/5, c = -7/10 Now,
a + b = -1/2 + 3/5 = (-5 + 6)/10 = 1/10
and (a + b) + c = 1/10 + (-7)/10
⇒ (a + b) + c = (1 – 7)/10 = -6/10 = -3/5
Also, b + c = 3/5 + (-7)/10
⇒ b + c = (6 – 7)/10 = -1/10
and, a + (b + c) = -1/2 + (-1)/10
⇒ a + (b + c) = (-5 – 1)/10= -6/10 = -3/5
Thus, (a + b) + c = a + (b + c) is true for Rational Numbers.
Similarly, for multiplication
a × b = -1/2 × 3/5 = -3/10
and, (a × b)× c = -3/10 × -7/10= -3× (-7)/100
⇒ (a × b)× c = 21/100
Also, b× c = 3/5 × (-7)/10 = -21/50
and, a × ( b × c ) = -1/2 × (-21)/50
⇒ a × ( b × c ) = 21/100
Thus, (a× b)× c = a × ( b × c ) is true for Rational Numbers.
Identity Property
Identity Property for Addition: For any rational number a, there exists a unique rational number 0 such that 0 + a = a = a + 0, where 0 is called the identity of the rational number under the addition operation.
Identity Property for Multiplication: For any rational number a, there exists a unique rational number 1 such that a × 1 = a = a × 1, where 1 is called the identity of the rational number under the multiplication operation.
Inverse Property
Additive Inverse property: For any rational number a, there exists a unique rational number -a such that a + (-a) = (-a) + a = 0, and -a is called the inverse of element a under the operation of addition. Also, 0 is the additive identity.
Multiplicative Inverse property: For any rational number b, there exists a unique rational number 1/b such that b × 1 / b = 1 / b × b = 1, and 1/b is called the inverse of the element b under the multiplication operation. Here, 1 is the multiplicative identity.
Example:
For a = -11/23 a + (-a) = -11/23 – (-11)/23 a + (-a) = -11/23 + 11/23 = (-11 + 11)/23 = 0
Similarly, (-a) + a = 0
Thus, 11/23 is the additive inverse of -11/23.
Now, for b = -17/29 b × 1/b = -17/29 × -29/17 = 1
Similarly, 1 / b × b = 29/17 × -17/29 = 1
Thus, -29 / 17 is the multiplicative inverse of -17/23.
Distributive Property
Distributive property for any two operations holds if one distributes over the other. For example, multiplicative is distributive over addition for the collection of rational numbers, for any three rational numbers a, b, and c the distributive law of multiplication of addition is
a × (b +c) = (a× b) + (a × c), and it is true for all the rational numbers.
Example:
For rational number a, b, c i.e., a = -7 / 9, b = 11 / 18 and c = -14 / 27
Now, b + c = 11/18 + (-14)/27 ⇒ b + c = 33/54 + (-28)/54 = (33 – 28)/54 = 5/54
and, a × ( b + c ) = -7/9 × 5/54 ⇒ a × ( b + c ) = (-7 × 5)/(9 × 54) = -35/486 . . .(1)
Also, a × b = -7/9 × 11/18 ⇒ a × b = (-7 × 11)/9 × 18 = -77/9 × 9 × 2
and a × c = (-7)/ 9 ×(-14)/27 ⇒ a × c = (7 × 14)/9 × 9 × 3 = 98/9 × 9 × 3
Now, (a × b) + (a × c) = (-77/9 × 9 × 2 ) + ( 98/9 × 9 × 3) ⇒ (a × b) + (a × c) = (-77 × 3 + 98 × 2)/9 × 9 × 2 × 3 ⇒ (a × b) + (a × c) = (-231 + 196)/486 = -35/486 . . .(2)
(1) and(2) shows that a × ( b + c ) = ( a × b ) + ( a × c ).
Hence, multiplication is distributive over addition for the collection Q of rational numbers.
With common denominator, rational number with greatest numerator is greatest.
⇒ -30/60 < -45/60 < 20/60 < 24/60 Thus, ascending order of given rational numbers is: -1/2 < -3/4 < 1/3 < 2/5
FAQs on Rational Numbers
Q1: What is the difference between rational and irrational numbers?
Answer:
A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating and non-recurring. An example of a rational number is 10/2, and an irrational number is a famous mathematical value Pi(π) which is equal to 3.141592653589…….
Q2: What are rational numbers?
Answer:
A rational number is a number that is in the form of p/q, where p and q are integers, and q is not equal to 0. Some examples of rational numbers include 1/3, 2/4, 1/5, 9/3, and so on.
Q3: Is 0 a rational number?
Answer:
Yes, 0 is a rational number because it is an integer that can be written in any form such as 0/1, 0/2, where b is a non-zero integer. It can be written in the form: p/q = 0/1. Hence, we conclude that 0 is a rational number.
Q4: Is Pi(π) a rational number?
Answer:
No, Pi (π) is not a rational number. It is an irrational number and its value equals 3.142857…
Q5: Are fractions rational numbers?
Answer:
Fractions are numbers that are represented in the form of (numerator/denominator) which is equivalent to p/q form so fractions are considered rational numbers. Example 3/4 is a fraction but is also a rational number.
Q6: Are all rational numbers integers?
Answer:
No, all rational numbers are not integers but the opposite is true. i.e. “all integers are rational numbers.” For example, 1/2 is a rational number but not an integer whereas -7 is an integer and is also a rational number.
Q7: Can rational numbers be negative?
Answer:
Yes, a rational number can be negative i.e. all negative number comes under rational numbers. Example -1.25 is a rational number.
Q8: Are all whole numbers rational numbers?
Answer:
Yes, all whole number are considered as rational numbers. For example 1 is a whole number and is also a rational number.
Q9: How many rational numbers are between 1/2 and 1/3?
Answer:
There are infinitly many rational numbers between any two rational number, thus there are infinitlyl many rational numbers between 1/2 and 1/3, some of those numbers are 11/24, 7/24, 19/48, 13/72, 3/8 etc.
Q10: By which symbol rational number denoted?
Answer:
Rational Numbers are denoted by “Q” in the mathematics.
. This is known as the recurring decimals expansion.
