Question

prove that 7-3√2 is an irrational number if it is given that √2 is an irrational number?​

Answer 1

Answer:

Let us assume 7-3 root 2 is rational,

Which is, 7-3root2 = p/q ( where p and q are 2 co-prime numbers having only 1 common factor, which is 1, and q is not equal to 0)

7-3root2 = p/q

-3root2 = p/q-7

-3root2 = p-7q/q

root2 = p-7q/-3q

We know that p-7q/-3q is rational, whereas root2 is irrational,

This is a contradiction to our assumption.

Therefore, 7-3root2 is irrational.

Hope It Helps.

:)

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Answer 2

$\bf \huge \underbrace{Answer} \\ \\ \sf \: given : - \\ \\ \sf \: \sqrt{2} \: \: is \: \: \: an \: \: irrational \: \: \: number \\ \\ now \\ \\ let \: \: us \: \: assume \: \: that \: \: 7 - 3 \sqrt{2} \: \: rational \: number \: \: and \: \: also \: \: co \: \: \: prime \: \: number \: \: \: a \: \: and \: \: b \: \: (b ≠0) \\ \\ \sf \implies \:7 - 3 \sqrt{2} = \frac{a}{b} \\ \\ \sf \implies \: - 3 \sqrt{2} = \frac{a - 7b}{b} \\ \\ \sf \implies \: \sqrt{2} = \frac{a - 7b}{ - 3b} \\ \\ \: \frac{a - 7b}{ - 3b} \: \: is \: \: a \: \: rational \: \: number. \\ \\ but \: \: in \: \: given \: \: \sqrt{2} \: \: is \: \: a \: \: irrational \: \: number \\ \\ so \: \: our \: \: assumption \: \: \: is \: \: wrong \: \: \\ \\ therefore \\ \\ 7 - 3 \sqrt{2} \: \: is \: \: a \: \: irrational \: \: number \\ \\ \bf \huge \blue{hence \: \: proved}$