Question

prove that 6 + root 2 is an irrational number

Answer 1

Answer:

6+√2 is irrational.

Step-by-step explanation:

Let us assume that 6+√2 is rational.

That is , we can find coprimes a and b (b≠0) such that

$6+\sqrt{2}=\frac{a}{b}$

$\implies \sqrt{2}=\frac{a}{b}-6$

$\implies \sqrt{2}=\frac{a-6b}{b}$

Since , a and b are integers , $\frac{a-6b}{b}$ is rational ,and so √2 is rational.

But this contradicts the fact that √2 is irrational.

So, we conclude that 6+√2 is irrational.

•••♪

Answer 2

Step-by-step explanation:

Let 6+root 2 is a rational number then we get that a and b two co-prime integers.

Such that 6+root 2=a/b where b not equal

root 2 = a/b - 6

root 2 =(6-a/b)

Since a and b are two integers.

Therefore (6-a/b)is a rational number and So root also is a rational number.

But it is contradiction to fact root 2 =(6-a/b) is rational number.

So we that 6+root 2 is an irrational number.