show that 4+3√2 is an irrational number
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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.✌️
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