prove that √2 is an irrational no.
Answers
Easy
Most common question
Lets first assume √2 to be rational.
√2 = a/b, where b is not equal to 0.
Here, a and b are co-primes whose HCF is 1.
√2 = a/b ( squaring both sides )… 2 = a^2/b^2
2b^2 = a^2 ….. Eq.1
Here, 2 divides a^2 also a ( bcz, If a prime number divides the square of a positive integer, then it divides the integer itself )
Now, let a = 2c ( squaring both sides )…
a^2 = 4c^2…Eq.2
Substituting Eq. 1 in Eq. 2,
2 b^2 = 4c^2
b^2 = 2c^2
2c^2 = b^2
2 divides b^2 as well as b.
Conclusion:
Here, a & b are divisible by 2 also. But our assumption that their HCF is 1 is being contradicted.
Therefore, our assumption that √2 is rational is wrong. Thus, it is irrational.
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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.