proof 1/√2 is irrational
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To prove:-
let 1/√2 is a rational number
so, we write it in the form of rational number
1/√2 = a/b
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 simplifya/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; ifa 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/b22=4k2/b22*b2=4k2b2=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/bwas 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.
let 1/√2 is a rational number
so, we write it in the form of rational number
1/√2 = a/b
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 simplifya/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; ifa 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/b22=4k2/b22*b2=4k2b2=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/bwas 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.
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