Question

Prove that 3√2-5 is irrational

Answer 1

let as assume 3√2-5 as rational..

where a and b are integers.
a/b=3√2-5
a+5b/b=3√2(L.C.M.)
a+5b/3b=√2

where a and b are integers, a+5b/3b is rational so, √2 is also rational
but the contradict fact was that √2 is rational 
so our assumption of √2 is rational is incorrect.

so, we conclude 3√2-5 as irrational..


hope its helps u..YO!!///!!@@


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Answer 2

Heya!!!

☜☆☞Here's your answer friend,

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Let 3√2 - 5 be a rational number,

therefore, 3√2 - 5 = a / b

.... (where a and b are coprime numbers and b ≠ 0)

==> 3√2 = a / b + 5

==> 3√2 = (a + 5b) /b

==> √2 = (a + 5b) / 3b

since a and b are integers ( a + 5b) / 3b is a rational number

there fore, √2 is a rational number

But this contradicts the fact that √2 is an irrational number.

Hence, our assumption proved wrong

==> √2 is an irrational number.

==> 3√2 - 5 is an irrational number.

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Hope it helps you : )