Prove that root 2 is an irrational number.

We prove this by contradiction. Assume it to be a rational number, so it can be expressed in the form of a/b where a & b are co-prime integers(Having 1 as only common factor). root 2 = a/b squaring both sides 2 = square of a/ square of b. 2 is an integer, whereas square of a/ square of b is a fraction, as a & b are co-prime integers so is square of a & square of b. Thus we arrive at the wrong conclusion. Hence our assumption is wrong. Thus square root of 2 is an irrational number. read less

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 = a^2/b^2, or a^2 = 2 · b^2. 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 = a^2/b^2, this is what we get: 2 = (2k)2/b2 2 = 4k^2/b^2 2*b^2 = 4k^2 b2 = 2k2 This means that b^2 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. read less

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. read less

Suppose, to the contrary, that Sqrt[2] were rational. Then Sqrt[2]=m/n for some integers m, n in lowest terms, i.e., m and n have no common factors. Then 2=m2/n2, which implies that m2=2n2. Hence m2 is even, which implies that m is even. Then m=2k for some integer k. So 2=(2k)2/n2, but then 2n2 = 4k2, or n2 = 2k2. So n2 is even. But this means that n must be even, because the square of an odd number cannot be even. We have just showed that both m and n are even, which contradicts the fact that m, n are in lowest terms. Thus our original assumption (that Sqrt[2] is rational) is false, so the Sqrt[2] must be irrational. read less

Nothing more to add here. the above details should have sufficed hour query. But if you have a doubt that why do we have to prove it by contradiction i have an argument for you. we dont have a direct definition for irrational number. its definition is : a number which is not a rational number. so the motive behind the proof should be to show 2^0.5 not a rational number. rational number : a number which can be expressed a p / q where p,q are non-zero interzers which have gcd = 1 read less