Question

3) Show that 5+√
7 is an irrational number​

Answer 1

Answer:

Let us assume to the contrary that 3√5-7 is a rational no.

Such That ,

3√5-7=p/q {where, p and q are integers having no common factors}.

3√5=p/q+7

3√5=.p+7q/q

√5= p+7q/3q

where , √5 and p+7q/3q are rational numbers.

But this contradicts the fact that √5 is irrational.

Therefore,3√5-7 is an irrational no.

MARK AS BRAINLIEST !!

Answer 2

let us assume that 5+√7 is an rational number

such that 5+√7=a/b (b≠0) . a and b are coprime integers

so,

5+√7=a/b

on squaring both side

(5+√7)² =a²/b²

(5)²+(√7)²+2×5×√7=a²/b²

25+7+10√7=a²/b²

32+10√7=a²/b²

10√7=a²/b²-32

10√7=(a²-32b²)/b²

√7=(a²-32b²)/7b²

hence, a and b are coprime integers

so, (a²-32b²)/7b² is rational number

then √7 is rational number

now, our assumption is wrong

so , 5+√7 is an irrational number