prove that \sqrt{ 2} is irrational?
Answers
Let us assume that √2 is a rational number.
then, as we know a rational number should be in the form of p/q
where p and q are co- prime number.
So,
√2 = p/q { where p and q are co- prime}
√2q = p
Now, by squaring both the side
we get,
(√2q)² = p²
2q² = p² ........ ( i )
So,
if 2 is the factor of p²
then, 2 is also a factor of p ..... ( ii )
=> Let p = 2m { where m is any integer }
squaring both sides
p² = (2m)²
p² = 4m²
putting the value of p² in equation ( i )
2q² = p²
2q² = 4m²
q² = 2m²
So,
if 2 is factor of q²
then, 2 is also factor of q
Since
2 is factor of p & q both
So, our assumption that p & q are co- prime is wrong
Hence √2 is an irrational number
Step-by-step explanation:
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.