Prove that is irrational and hence prove that is irrational.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that $\sqrt{2}$ is irrational and hence prove that $\frac{5 - 3 \sqrt{2}}{7}$ is irrational.

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Let us assume that $\sqrt{2}$ is a rational. Then there exist co-prime positive integers a and b such that,

\sqrt{2} = \frac{a}{b}

⟹ a = b \sqrt{2}

Squaring on both sides, we get

a 2 = 2b 2

Therefore, a² is divisible by 2 and hence a is also divisible by 2

So, we can write a = 2p, for some integer p

substituting for a, we get

4p 2 = 2b 2 \Rightarrow b 2 2p 2

This means, b² is divisible by 2 and so, b is also divisible by 2.

Therefore, a and b have at least one common factor, i.e, 2

But, this contradicts the fact that a and b are co-primes.

Thus, our supposition is wrong.

Hence,

\sqrt{2}

is an irrational.

\sqrt{2}

is an irrational ,

5 - 3 \sqrt{2}

is an irrational

And

\frac{5 - 3 \sqrt{2}}{7}

is also an irrational.