Question

Prove That 7-√5 is an irrational number

Answer 1

hey., mate,

Given : 7 − √5 LET US ASSUME THAT 7 − √5 IS RATIONAL NUMBER ∴ 7 − √5 = a/b (a,b ∈ z , where b ≠ 0) −√5 = a/b − 7 √5 = 7 − a/b IF a,b ARE INTEGERS THEN 7 − a/b IS RATIONAL NUMBER. SO √5 IS ALSO A RATIONAL NUMBER. BUT THIS CONTRADICTS THE FACT THAT √2 IS IRRATIONAL. ∴ WE CONCLUDE THAT 7 − √5 IS IRRATIONAL NUMBER.Read more on Sarthaks.com - https://www.sarthaks.com/2956/prove-that-7-root-5-is-an-irrational-number

HOPE THIS MAY HELP YOU

Answer 2

let assume7- √5 is a rational number.

so √5=p/q where q is not equal to zero

$7 - \sqrt{5} = \frac{p}{q} \\ 7q - \sqrt{5} q = p \\ 7q - p = \sqrt{5} q \\ \frac{7q - p}{q} = \sqrt{5}$

hence LHS =RHS so we know that √5 is irrational number so7-√5 is also an irrational number.

Hence it contradict our assumption.

hope it helps....

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