Question

PROVING METHOD OF CONTRADICTION WRONG:
an excerpt from a book on method of contradiction:
A proof that the square root of 2 is irrational
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where
a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that
can obviously be done with any fraction. Notice that in order for a/b to be in
simplest terms, both of a and b cannot be even. One or both must be odd.
Otherwise, we could simplify a/b further.

From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the
square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it
can't be odd; if a itself was odd, then a · a would be odd too. Odd number
times odd number is always odd. Check it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times some other whole
number. In symbols, a = 2k where k is this other number. We don't need to
know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we
get:

2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even.
And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process
assuming that a/b was simplified to lowest terms, and now it turns out that a
and b both would be even. We ended at a contradiction; thus our original
assumption (that √2 is rational) is not correct. Therefore √2 cannot be
rational.

However,if
√(4)=a/b
consider that a & b are coprimes
a^2=4b^2
a is divisible by 4.
let k=4a
16k^2=4b^2
4k^2=b^2
b is divisible by 4

thus,a & b are not coprimes as they are divisible by 4

so √4 is irrational≠true

so is the method of contradiction wrong??

Answer 1

yes our contradiction is wrong a and b are integer