 # CBSE / NCERT Maths Solutions Class 10 – Real Numbers – Irrational Theorems ## CBSE / NCERT Maths Solutions Class 10 – Real Numbers – Irrational Theorems

In this article, we will discuss Irrational Numbers from Class 10 Maths Chapter 1 Real Numbers.

Irrational Numbers

Irrational numbers which when expressed in decimal forms are expressible as non-terminating and nonrepeating decimals are known as irrational numbers.

0.1010010001…is a non-terminating and non-repeating decimal.

0.2020020002

0.12112111211112 is irrational.

0.232322223 is irrational.

22/7 is irrational.

Some Results on Irrationals

Theorem 1

Let P be a prime number and a be a positive integer. If p divides then show that p divides a.

Proof: – Let p be a prime number and a be a positive integer such that p divides

We all are aware that every positive integer can be expressed as the product of primes.

Let a = p1 , p2 ……. pn where   , ….. , are primes p1, p2 ……. pn
a2 = (p1, p2 ……. pn) (p1, p2 ……. pn)

a2 = (p12, p22 ……. pn2)

P divides a2

P is one of  p1, p2 ……. pn

P divides a

Learn how to use Theorem 1 in solving questions, click CBSE Maths Class 10th for demos.

Theorem 2

Prove that √2 is irrational.

Proof: – Let 2 be rational and let its simplest form be a/b

Then a an b are integers having no common factors other than 1 and b is not equal 0

√2=a/b

2=a²/b²

2b²=a²

2 divides a

2 is prime and divides a²

Let a=2c for some integer c

Putting a=2c we get

2b²=4c²

b²=2c²

2 divide b²

2 divide b

2 is prime and divides b²

2 divide b

2 is a common factor of a and b

But this contradicts the fact that a and b have no common factor other than 1

The contradiction arises by assuming that √2 is rational

Hence 2 is irrational

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Theorem 3

Prove that √3 is irrational

If possible √3 be rational and let its simplest form be a/b

Then a and b are integers having no common factors other than 1 and b is not equal 0

√3=a/b

3=a²/b² on squaring both sides

3b²=a²

3 divides a

3 is prime and 3 divides a²

3 divides a

Let a=3c for some integer c

Putting a =3c we get

3b²=9c²

b²=3c²

3 divide b²

3 divide b

3 is prime and 3 divide b². Thus, 3 is a common factor of a and b

But this contradicts the fact that a and b have no common factors other than 1.

The contradiction arises by assuming √3 is rational.

Hence 3 is irrational.

In the same manner, we can prove that each of the numbers √5, √7 is irrational.

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May 1, 2018

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