Pop Video: Proving that the Square Root of Two is Irrational | Nagwa Pop Video: Proving that the Square Root of Two is Irrational | Nagwa

# Pop Video: Proving that the Square Root of Two is Irrational

In this video we learn how to use the technique of proof by contradiction to prove that the square root of two is an irrational number. This means that the square root of two cannot be expressed as a fraction with integers as numerator and denominator.

12:50

### Video Transcript

In this video, we’re gonna run through a mathematical proof that the square root of two is an irrational number, which means that it’s a number that can’t be written as a fraction in which the numerators and denominators are both whole numbers.

Now that, in turn, means that if you try to write it out in decimal format, you’d have to go on working out decimal places forever, but more of that in a minute.

First, let’s talk about some ancient Greek philosophers, Pythagoras and Hippasus. I should think that, pretty well, all of you will have heard of Pythagoras because of the Pythagorean theorem, or Pythagoras’s theorem like some people call it.

It states that, in a right-angled triangle, the square of the length of the longest side is equal to the sum of the squares of the lengths of the shorter sides. You might know this better as 𝑎 squared plus 𝑏 squared equals 𝑐 squared. And it seems pretty easy because we learn about it quite early in our school careers. But it was a big deal back in Pythagoras’s day, about 2500 years ago.

The general agreement though that the Pythagorean theorem wasn’t written by Pythagoras, even if it did become one of his best-known greatest hits. Now there are lots of stories, myths, and legends floating around about Pythagoras. And whatever you say about him, someone else will probably say something different.

The problem is that none of his writings have survived the passage of time. And many of the things that were written about him by other people are contradictory. He seems to have been a bit of a controversial figure. He called himself a philosopher, a lover of wisdom. And he started a movement of followers called the Pythagoreans. They were very secretive. And there were many references to their beliefs in the mystical powers and purity of numbers.

One of Pythagoras’s followers was called Hippasus, although he didn’t entirely clear whether he followed Pythagoras himself or whether he was simply a follower of the Pythagorean movement. And maybe he wasn’t even born during Pythagoras‘s lifetime. But anyway, from ancient artwork and drawings that survived, it does seem likely that one thing they had in common was that they both had beards. These beards were very fashionable at the time. A few people would contradict this.

Okay, to bring this rambling tale to a point of focus, there’s a story that Hippasus managed to develop a proof that some numbers are irrational, which really upsettingly contradicted the Pythagorean police and the divine nature of numbers. There are reports that Hippasus drowned at sea a short while later.

Some people say he just picked a very bad time indeed to reveal the truth of irrational numbers to his Pythagorean friends, while out at sea on a ship. And they acted swiftly to cover it up. But this doesn’t quite hold water so to speak, because if it was true, how would we know about it?

Now Hippasus apparently showed how constructing a dodecahedron inside a sphere leads to the need for irrational numbers. But an easier method would have been to use a right-angled isosceles triangle with two sides having a length of one unit and then the Pythagorean theorem to show that, in this case, the length of the longest side would be equal to the square root of two units.

So we know that the need for a simple situation involving the square root of two arises. How can we now show that this is an irrational number? Well, first, let’s just make sure we know what rational and irrational numbers are.

A half can be called a ratio of one to two. It consists of a fraction with a whole number or integer on the numerator, one in this case, and another whole number or integer on the denominator, two in this case. That’s basically all a rational number is, a fraction with integers on the top and the bottom.

Remember, if the numerator and denominator both have a common factor, then we can simplify the fraction by dividing both by the same factor and getting an equivalent fraction. For example, two over four is a rational number. But we can divide the top and bottom by numbers by two to get the equivalent fraction, a half. A half is a simpler form of two-quarters. And it’s also a rational number.

Now we can take any integer we like as the numerator and any other integer we like as the denominator to make a rational number. Now let’s take a look at a couple of rational numbers. One-third and two-thirds, is there another rational number in between them in value? Well, one and a half thirds doesn’t really count because one and a half isn’t an integer.

But an equivalent fraction to one-third is two-sixths, just double the numerator and denominator. And an equivalent fraction to two-thirds is four-sixths, again just double the numerator and denominator. So instead of one-third and two-thirds, we’ve got two-sixths and four-sixths. And it sort of leaps off the page that three-sixths would be between them. And we could simplify that to a half by dividing the numerator and denominator by three.

Now we can apply this technique to any two rational numbers to find another rational number between them. And we can go on forever, getting smaller and smaller differences. So it begins to look like you ought to be able to make any number rational. So what’s an irrational number then?

Well, it’s a number that we can’t represent exactly by a fraction with integers as numerators and denominators. For example, we use the Pythagorean theorem to show how there’s a value called a square root of two. Now let’s play around with the concept a little bit to see if we can prove that you can’t find a pair of integers for numerators and denominators to represent that value.

Let’s start by assuming, in fact, that the square root of two is rational and that there are two integers — let’s call them 𝑎 and 𝑏 — that we can use as the numerator and denominator to represent that value. So we’ve got the square root of two is equal to 𝑎 over 𝑏, where 𝑎 and 𝑏 are integers and 𝑏 is not equal to zero because something divided by zero is undefined.

Let’s also choose to pick 𝑎 and 𝑏 so that they represent a fully simplified fraction version of root two. Obviously, there’ll be a family of equivalent fractions: two 𝑎 over two 𝑏, three 𝑎 over three 𝑏, and so on. But we’re choosing 𝑎 and 𝑏 to make the fully simplified fraction version so that they don’t have any common factors.

And there’s another implication of this. If 𝑎 is an even number, then 𝑏 must be odd. And if 𝑏 is an even number, then 𝑎 must be odd. If they were both even, then they’ll both be multiples of two. And two will be a common factor. So we’ll be able to cancel and simplify the fraction. But we chose 𝑎 and 𝑏 carefully so that they didn’t have any common factors.

Okay, so we’ve got the square root of two is equal to 𝑎 over 𝑏. Now let’s square both sides of the equation. And that gives us two is equal to 𝑎 squared over 𝑏 squared. Now I can multiply both sides of my equation by 𝑏 squared so that the 𝑏 squareds cancel on the right, which gives us that two 𝑏 squared is equal to 𝑎 squared.

However, on the left-hand side, 𝑏 remember is an integer. So 𝑏 times 𝑏, 𝑏 squared, that’s an integer times an integer that must also be an integer. So the left-hand side is two times a whole number. And a whole number which is a multiple of two is an even number.

Now over on the right-hand side, we’ve got 𝑎, an integer, times itself. So we’ve got a whole number times a whole number. And the only way that you can get an even result when you multiply two whole numbers together is if one of those whole numbers was even. And since we’re actually talking about 𝑎 times itself, then 𝑎 must be an even number.

Okay, let’s run through that logic in a little bit more detail. We can say that an even number is just a whole number multiple of two. So let’s pick a letter to represent any whole number. Let’s say 𝑚. Then we can say that two 𝑚 is an even number.

You tell me the even number you want and I’ll pick a suitable value for 𝑚 to generate that even number. It’ll just be half the value of the even number you want. You want eight, I’ll pick 𝑚 equals four. So two 𝑚 is the even number eight in this case. Two 𝑚 is just an expression representing a number that we know must be even.

Now we can represent another even number by assigning the letter 𝑛 to represent another whole number. And then two 𝑛 must be another even number. Now let’s multiply our two even numbers together, two 𝑚 times two 𝑛. And because multiplication is associative, we can write that as two times 𝑚 times two times 𝑛.

And since two 𝑚 and 𝑛 are all whole numbers, we know that 𝑚 times two times 𝑛 will also be a whole number. And that means that two times 𝑚 times two times 𝑛 is two times a whole number, which must be an even number. So if we multiply any two even numbers together, we definitely get a result that’s even.

Now odds and evens alternate throughout the whole numbers. One is odd, two is even, three is odd, four is even, five is odd, six even, and so on forever. And that means that because two 𝑚 is an even number, then two 𝑚 plus one must be the next odd number after it. Likewise, two 𝑛 is an even number. So two 𝑛 plus one must be the next odd number.

So let’s look at other combinations of multiplying odd and even numbers together to see if we can get an even number result. For example, if we wanted to multiply an even number by an odd number, we can do two 𝑚 times two 𝑛 plus one. And again, because of associativity, we can write that as two times 𝑚 times two times 𝑛 plus one.

And again, we’ve got integers inside the parentheses there. So we’ve got two times an integer. So even times odd also gives us an even number. And that works the other way round too. If we had an odd number times an even number, we’d also get an even number.

Lastly then, let’s try multiplying an odd number by an odd number. Then multiplying each term in the first parentheses by each term in the second parentheses gives us two 𝑚 times two 𝑛 plus two 𝑚 times one plus one times two 𝑛 plus one times one, which simplifies to four 𝑚𝑛 plus two 𝑚 plus two 𝑛 plus one.

And if we factor out a two from these first three terms here, we get two times two 𝑚𝑛 plus 𝑚 plus 𝑛 all plus one. Now two, 𝑚, and 𝑛 are all integers. And we’re multiplying and adding those together. So the contents of those parentheses there are gonna be an integer. This gives us two times an integer, which is an even number. So the result is gonna be an even number plus one, which is an odd number. And that means if I multiply any two odd numbers together, then the result is another odd number.

So back to our problem, we had two 𝑏 squared equals 𝑎 squared. Now we said that the left-hand side, because 𝑏 is an integer, must be an even number. And the right-hand side is equal to 𝑎 squared, which is something times itself. So we’re either dealing with an even number times an even number or an odd number times an odd number. Now the only way we can get an even result is if 𝑎 was even. So this is definitely true.

Right, remember that I said if you want a specific even number, I can halve it and express your even number as two times half of that number. Well, let’s do the same thing for the even number 𝑎. Let’s call half of 𝑎 𝑐. That means that 𝑐 is equal to a half of 𝑎.

In other words, two 𝑐 is equal to 𝑎. And we can replace 𝑎 with two 𝑐 in our equation, which means that two 𝑏 squared is equal to two 𝑐 all squared and that two 𝑐 all squared means two 𝑐 times two 𝑐. So we now know that two 𝑏 squared is equal to four 𝑐 squared.

Now I can divide both sides by two to cancel off the twos here. And I get two and one over here. In other words, 𝑏 squared is equal to two times 𝑐 squared. Then over on the right-hand side, we said 𝑐 is an integer. So 𝑐 squared, an integer times an integer, is also an integer. And this expression here, two times an integer, must give us an even result.

Then using the same logic we used over here to prove that 𝑎 must be even, we can say that 𝑏 must also be even. But wait a minute! We said right near the beginning that if 𝑎 is an even number, then 𝑏 must be odd. And if 𝑏 is an even number, then 𝑎 must be odd. Now we’ve just shown that 𝑎 must be even and 𝑏 must be even. This is a contradiction.

We’ve shown that 𝑏 must be both odd and even at the same time. This must mean that our original assumption was wrong. We assumed that there are two integers 𝑎 and 𝑏 that can be used as the numerator and denominator in a fraction in its simplest form representing the value of the square root of two.

But that assumption leads us to two mutually exclusive conclusions. That is, 𝑏 is an even number and 𝑏 is an odd number. So the assumption must have been wrong. There aren’t two integers 𝑎 and 𝑏 that can be used as a numerator and denominator in a fraction in its simplest form representing the value of the square root of two. We call this sort of proof “proof by contradiction.” Rather than proving something is true in all cases, we’ve proved that assuming it’s true leads us to a nonsense situation. So it can’t be true. It’s a pretty powerful technique.

Now irrational numbers have been known about for a very long time. And people don’t tend to get as upset as the Pythagoreans did when they find out about them these days. But if you’re gonna show this proof of the irrationality of the square root of two to anyone, maybe it might be as well to make sure you’re on dry land when you do it, just in case.

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