Question

prove that 7+3√2 is an irratonal number​

Answer 1

Let 7 + 3√2 be an rational number where

7+3√2 = a/b [ a and b are coprime and b is not equal to zero]

3√2= a/b-7

3√2 =( a-7b) /b

√2 = (a-7b) /3b .....(i)

Now ,from equation (i) ,we get that √2 is rational but we know that √2 is irrational.

So actually 7 + 3√2 is irrational not rational. thus our assumption is wrong.

Hence , 7+3√2 is an irrational number

Answer 2

Answer:

Let's assume that 7+3√2 is a rational no.

let p/q = 7+ 3√2 -(where people and q are coprime and q ≠0)

=p/q -7 = 3√2

=p -7q/3q =√2

here we know that√2 is irrational no.

(P -7q/3q) is a factor of √2.

our assumption is wrong.

this contradicts that 7+ 3√2 is irrational no.

hence, proved.

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