Question

Show that 7-5√2 is irrational​

Answer 1

Let 7-5√2 be rational

7-5√2 = p/q ; where p and q are co prime and q ≠ 0

√2 = - p/5q +7/5

In LHS there is Irrational no. And in RHS there is rational no. Thus this is no possible. Thus contradiction arise due to our wrong assumption.

Hence our assumption is wrong

7-5√2 is Irrational no.

Hence proved

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Answer 2

Step-by-step explanation:

Let us assume that 7-5√2 is a rational number.

We can express this as;

$7-5 \sqrt{2} = \frac{p}{q}$

where p and q are integers and q ≠ 0

$7 - \frac{p}{q} = 5 \sqrt{2}$

$\frac{7q - p}{q} = 5 \sqrt{2}$

$\frac{7q - p}{5q} = \sqrt{2}$

$where \: \frac{7q - p}{5q} \: is \: rational.$

$but \: \sqrt{2} \: is \: irrational.$

This contradicts the fact that √2 is irrational.

Hence our assumption is wrong.

Hence we conclude that 7-5√2 is irrational.