Question

prove that 4 - 3 root 2 is irrational if it is given that root 2 is an irrational no .


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Answer 1

Answer:

Let us assume that $4-3\sqrt[]{2}$  is rational

Therefore, $4-3\sqrt[]{2}$ = $\frac{a}{b}$, where a and b are coprime.

$4-\frac{a}{b} = 3\sqrt[]{2}$

$\frac{4b-a}{b} = 3\sqrt[]{2}$

Therefore, $\sqrt[]{2} = \frac{4b-a}{3b}$

LHS is irrational, RHS is rational

This contradicts the fact that $\sqrt[]{2}$ is irrational

This contradiction has arisen because our assumption is wrong

Thus, $4-3\sqrt[]{2}$ is irrational

Hope this helps :)

Answer 2

Answer:

irrational number

Step-by-step explanation:

Let us assume that 4-3√2 is rational number.

So we can write 4-3√2 as a/b where a and b are co primes and b is not equal to 0.

4-3√2 = a/b.

-3√2 = a/b-4.

-3√2= a-4b/b

√2= a-4b/-3b

√2 = -a-4b/3b.

Here √2 is an irrational number.

But a-4b/-3b or -a-4b/3b is rational number.

Therefore it is a contradiction to our assumption that 4-3√2 is rational number.

Thus,4-3√2 is irrational number...

Hope this helps.