Question

How to prove that 4-3√2 is irrational

Answer 1

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 you.

Answer 2

Answer:

Hey, mate here's your answer

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...

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