Question

prove that √7 is irrational​

Answer 1

Answer:

when the number have square root on it's head is root. √7 is the irrational number

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Answer 2

Answer:

Mark as Brainlist answer.

Step-by-step explanation:

Let us assume that

7 is rational. Then, there exist co-prime positive integers a and b such that

√7=a/b

⟹a=b√7

Squaring on both sides, we get

a^2=7b^2

Therefore, a^2 is divisible by 7 and hence, a is also divisible by 7

so, we can write a=7p, for some integer p.

Substituting for a, we get 49p^2=7b^2⟹b^2=7p^2.

This means, b^2 is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

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

Thus, our supposition is wrong.

Hence,

√7 is irrational.