Question

Prove the following as irrational numbers 4-3√2

Answer 1

Answer:

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...

Answer 2

Step-by-step explanation:

it is irrational number

Let's 4 - 3√2 is a rational number equal to fraction a/b where A and B are rational number and b is not equal to zero.

4 - 3√2 = a/b

- 3√2 = a/b - 4

√2 = a/b - 4/3

we know that root 2 is irrational number and also so we have taken a and b as a rational number but if we subtract 4 upon 3 from A and B will get a rational number so the equation get strong and it gets an equal which proves that 4 - 3√2 is irrational .

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