If a Square Root is Not an Integer, What is It?
A Most Merry and Illustrated Irrational Proof
Hippasus was supposed to be the first.
There's one question that continually keeps people up at night. They worry about it, they fret about it, they keep wringin' their hand's an cryin'. And when they wake up in the mornin' the blues keep falling like midnight rain.
You know the question. You too have worried about it every day of your life. But you deny you are worried. So we will state the question clearly and openly as the first step in bringing help and aid to those in need.
How, just how do we really know that irrational numbers are indeed irrational?
Now when we say a number is irrational, we don't mean that it votes for political parties who will work against their interests or go to casinos and bet using double up systems on games with a high house percentage and a narrow range of lower and upper betting limits. No, no no, no. When we say a number is an irrational number we mean it cannot be expressed as a ratio of two integers.
The Greeks, though - particularly the Pythagoreans - thought all numbers were rational. But Hippasus of Metapontum (wherever that is) was able to prove that there were indeed numbers that were not expressible as the ratio of two integers.
Now when most people think of rational numbers, they think of fractions such as:
1/2, 2/3, 3/4, 100/101
On the other hand, whole number are rational too since you can express them as ratios of integers.
2/1, 100/10, 5000/500, 9^{99}/9
And of course there are "improper fractions":
3/2, 5/3, 355/113, 100001/100000
The problem in proving numbers are irrational is there are lots of different kinds of irrational numbers. For instance, the famous number π is irrational but it is also transcendental, as is the base of natural logarithms, the number e. A transcendental number means it is not the solution of polynomial equations where the coefficients are rational - the ax^{2} + bx + c equations that drove you nuts when in middle school. So there is no one proof to prove all irrational numbers are irrational.
But each journey must begin with a single step. So we'll start off with the simple irrational numbers.
Without doubt the simplest irrational number is the square root of 2, √ 2 . We know the number exits, as did Hippasus, since it's the length of the hypotenuse of a right triangle with the other two sides equal length 1 (you pick the units). Pythagoras in his famous theorem proved that the square of the hypotenuse was equal to the sum of the squares of the other two sides. So the right triangle with unit sides must have a hypotenuse equal to √ 2 .
Now it's not that people doubt √ 2 is irrational. But their big worry is how to prove it. Then to their great distress they find that friends, relatives, and co-workers can't help them out and seem surprisingly indifferent to the whole question. But worry no more! CooperToons as a public service will provide the proof. No, no, don't thank me. Glad to do it.
The proof that numbers are irrational is a type of proof that disturbs some people. That's because it is an indirect proof or proof by contradiction. You start off with the opposite of what you want to prove. Then you reason around a bit and hope you find something that contradicts some accepted mathematical fact. Do that, logicians tell us, and that proves the opposite of what you assumed - that is, what you wanted to prove in the first place - is indeed correct.
As we said, some people don't like indirect proofs. The concept itself is foreign to many - assuming the opposite of something to prove it's true. Then when you arrive at a contradiction, the contradiction seems to be irrelevant to the whole point you're trying to prove. That, though, is the beauty of it. Any contradiction of a known mathematical fact that you legitimately arrive at is good enough to establish the proof. So rest assured, proof by contradiction is a perfectly correct way to prove something. As an American president once said, "Trust me."
All right. So we assume the opposite of what we want. That is, we start off by saying that the square root of 2 is rational and so is equal to the ratio of two integers, a and b.
√ 2 = a / b
Well, if that is true, then 2 must the be ratio of two squares
2 = a^{2} / b^{2}
Or a^{2} is twice the size of b^{2}
a^{2} = 2b^{2}
Now since b is an integer b^{2} is also an integer. So if a^{2} is equal to 2 times another integer, it must be even.
Now if a^{2} is even, then it, too, must be expressible as 2 times another integer (which may be odd).
a^{2} = 2n
Therefore it follows that
b^{2} | = | a^{2} / 2 |
= | (2n)^{2} / 2 | |
= | 4n^{2} / 2 | |
= | 2n^{2} |
Hmmmmmm. Since n is an integer then n^{2} is an integer and so 2n^{2} must be an even number. That means b^{2} is also an even number.
At this point we've shown that both a^{2} and b^{2} are even if our original premise - √ 2 = b/a - is correct. So far so good.
Now we need to take - not the pause that refreshes - but the pause that lets us slip in another proof. We know that a^{2} and b^{2} are even numbers. But to go on with our proof we need to show that if a^{2} and b^{2} are even then both a and b are also even.
If you try some examples you'll find that if you find a square is an even integer and has an integer square root, then the square root is indeed an even number. But no matter how many examples we find, we have not proven if a^{2} and b^{2} are even, then a and b are even. To do that we need to try an oddball type of proof that is stranger a proof by contradiction. That is a proof by contrapositive.
So what is a proof by contrapositive? Well, an example is in order. Suppose I make the statement:
If that bird is a crow, then it is black.
Now as long as we ignore the possibility of albino crows and don't split hairs on what we mean by black, this statement is true. If a bird is a crow, then it is indeed black.
Now you can see that another way to say this is that:
If that bird is not black, then it is not a crow.
This, then, is the contrapositive of the first statement. The point, though, is the two statements are equivalent. And if you can prove one of them, then the other one is automatically proven as well.
OK. So we want to prove:
If n^{2} is even, then n is even.
It's hard to see how to prove this. There's really no easy route from our supposition - that n^{2} is even - to our conclusion - that n is even.
But what about the contrapositive
If n is not even, then n^{2} is not even.
Now since a number must be even or odd but cannot be both, we can rewrite our contrapositive as:
If n is odd, then n^{2} is odd
So are you saying if I can prove "If n is odd, then n^{2} is odd", then it is the same thing as proving "If n^{2} is even, then n is even".
Yep that's exactly what we are saying. And if that's a bit hard to believe, as an American president said ...
OK. So we'll proceed with our proof. Now if n is odd, then n = 2m + 1 where m is any integer.
So if n = 2m + 1, then
n^{2} | = | (2m + 1)^{2} |
= | (4m^{2} + 4m + 1) | |
= | 2(2m^{2} + m) + 1 |
Now if m is an integer, then 2m^{2} + m is an integer. That means that 2(2m^{2} + m) is an even number and so 2(2m^{2} + m) + 1 is an odd number. But 2(2m^{2} + m) + 1 was the square of an odd number.
So we've proven if a number is odd, then the square of the number is odd.
But remember this is the contrapositive of "If the square of a number is not odd, then the number is not odd".
In other words we've just proven "If the square of a number is even, then the number is also even".
Now we know that a^{2} and b^{2} are even if √ 2 = b/a. And with this last proof we now know that if √ 2 = b/a, then both a and b are also even.
However, two even numbers share a common factor, namely 2. In other words, we now conclude that if √ 2 is rational, it can only be expressible as the ratio of two numbers which must always share a common factor.
But hold on there! There is no rational number that is expressible only as two integers which share a common factor. Every rational number can always be expressed in reduced form, that is as the ratio of two integers which have no common factors. So if √ 2 is a rational number, then we can prove something that is clearly false. That is, we've found our contradiction.
Therefore we must conclude our assumption, that √ 2 is rational, cannot be true. Therefore $\sqrt{2}is\; an\; irrational\; number.$
Case closed. Q. E. D. Quod erat demonstrandum.
What? Something still bothering you?
Well, you're not alone if you have still feelings of misgiving. You may, in fact, be worrying about one of the commonly expressed objections.
You say we have a number that is not reducible to the least common factors? Well, then, we can just reduce them. That is, since we know that
√ 2 = 2m / 2n
then we just factor away the 2's.
√ 2 = 2m / 2n = m/n
So √ 2 is a rational number after all, nicht wahr?
Nope, sorry. Remember, the proof showed that if √ 2 is expressible as the ratio of two integers, the two numbers must always be even. Factor away two 2's and you can still prove the remaining integers - m and n - still must both be even. So no matter what, if √ 2 is rational, it is a rational number that cannot exist.
So √ 2 is an irrational number.
Now we can all sleep at night.
References
Prove It: The Art of Mathematical Argument, Bruce H. Edwards, University of Florida, The Great Courses, (The Teaching Company). The internet is replete with proofs but this course is a good introduction to mathematical proofs and indeed has the proof that √ 2 is irrational.
Most proofs of this theorem start out with the assumption that √ 2 = a/b and that a/b is in its reduced form. This approach seems to cause some confusion amongst readers probably because in this proof by contradiction we are starting off not with a single assumption but with two assumptions. So the alternative is what we have done here. We simply prove that if √ 2 = a/b then a and b must both always be even and so a/b cannot be written in reduced form. Only then do we point out such a rational number can never exist. The two approaches are equivalent.
"When is a square root irrational?", http://mathschallenge.net/library/number/rational_powers. The generalized proof.