Pop Video: Proving That 1 = 2 Using Basic Algebra

Video Transcript

In this video, we’re gonna be using some basic algebra to prove that one is equal to two. Well, maybe.

First of all let’s define two variables, 𝑝 and 𝑞. And we’ll say that the value of 𝑝 is equal to the value of 𝑞. Now that gives us the equation 𝑝 is equal to 𝑞. Now let’s take that equation and multiply both sides by 𝑝. So 𝑝 times 𝑝 is equal to 𝑝 times 𝑞 or simply 𝑝𝑝 is equal to 𝑝𝑞. But wait! 𝑝𝑝 just means 𝑝 times 𝑝 or 𝑝 squared, so we can say 𝑝 squared is equal to 𝑝 times 𝑞 or 𝑝𝑞.

And now let’s subtract 𝑞 squared from each side of that equation. So we’ve done the same thing to both sides of that equation, so it’s still an equation. And 𝑝 squared minus 𝑞 squared is equal to 𝑝𝑞 minus 𝑞 squared. Now when we’ve got a square thing minus another square thing, we call this the difference of two squares. And we can factorize that expression like this: 𝑝 squared minus 𝑞 squared is equal to 𝑝 minus 𝑞 times 𝑝 plus 𝑞.

So let’s just check that: positive 𝑝 times positive 𝑝 is positive 𝑝 squared; positive 𝑝 times positive 𝑞 is positive 𝑝𝑞; negative 𝑞 times positive 𝑝 is negative 𝑞𝑝; and negative 𝑞 times positive 𝑞 is negative 𝑞 squared. Now 𝑝𝑞 means 𝑝 times 𝑞, and 𝑞𝑝 means 𝑞 times 𝑝. But multiplication is commutative, and that means that it doesn’t matter in which order you multiply things together in. So 𝑝 times 𝑞 gives you the same result as 𝑞 times 𝑝, and that means that I can re-express 𝑞𝑝 as 𝑝𝑞.

So our expression now reads 𝑝 squared plus 𝑝𝑞 minus 𝑝𝑞 minus 𝑞 squared. And 𝑝𝑞 takeaway 𝑝𝑞 is nothing. And in multiplying out 𝑝 minus 𝑞 times 𝑝 plus 𝑞, I’ve proved that these two things are in fact equal. And that means I can write out the left-hand side of my equation like this. So I’ve now got 𝑝 minus 𝑞 times 𝑝 plus 𝑞 is equal to 𝑝𝑞 minus 𝑞 squared.

So I’ve factorized the left-hand side of our equation. But over on the right, we can see that both terms have a common factor of 𝑞 as well, so I can factorize that side too. So we’ve now got 𝑝 minus 𝑞 times 𝑝 plus 𝑞 is equal to 𝑞 times 𝑝 minus 𝑞. Now you’ll notice we’ve got 𝑝 minus 𝑞 on both sides of our equation, so if I divide both sides by 𝑝 minus 𝑞, I will have 𝑝 minus 𝑞 in the numerator and the denominator on both sides and I can cancel them out.

𝑝 minus 𝑞 divided by 𝑝 minus 𝑞 on the left-hand side is one and likewise on the right-hand side. So on the left-hand side, I’ve got one times 𝑝 plus 𝑞 over one, which is just 𝑝 plus 𝑞. And on the-right hand side, I’ve got 𝑞 times one over one, which is just 𝑞. And that leaves me with 𝑝 plus 𝑞 is equal to 𝑞.

Now remember, back at the beginning we said that 𝑝 was equal to 𝑞, and that means I can replace 𝑝 with 𝑞 in my equation, which gives us 𝑞 plus 𝑞 is equal to 𝑞. Well 𝑞 plus 𝑞 is just equal to two 𝑞. So now two 𝑞 is equal to 𝑞. Now I can Divide both sides by 𝑞; and on the left, 𝑞 divided by 𝑞 is one; and on the right, likewise, 𝑞 divided by 𝑞 is one. And that leaves us with two times one over one on the left-hand side, which is just two, and one divided by one on the right-hand side, which is just one.

So there we have it! Two is equal to one, or, indeed, one is equal to two. Well there it is; we’ve broken maths! Not much point in carrying on! Okay, pause the video, have a look through line by line, and see if you can find any errors in our logic.

Well, obviously one isn’t equal to two. If it was, I think we’d have heard it on the news by now. So let’s go through our working out line by line and see if we can work out what’s happened. Well defining two variables 𝑝 and 𝑞 and letting them be equal, there’s no problem with that. And it’s perfectly okay to multiply both sides of our equation by the same thing 𝑝. And yes, 𝑝 squared is equal to 𝑝 times 𝑝, so that’s correct.

Also subtracting the same thing from each side of our equation does keep it equal, so that’s alright too. Now we looked at the difference of two squares factorization in some detail earlier on, and that’s perfectly okay as well. And factorizing out the common factor of 𝑞 on the right-hand side, that’s perfectly okay.

Now in moving from step six to step seven, we divided both sides of the equation by 𝑝 minus 𝑞. Well this might normally be okay, but we did say at the beginning that 𝑝 was equal to 𝑞. So 𝑝 minus 𝑞 is equal to something minus itself; that’s zero. So we are dividing both sides of our equation by zero. Now that is a problem. For example, what’s one divided by zero? Well it doesn’t matter how many times you add zero to itself, you’re never going to reach one.

Anything divided by zero is undefined. So when you divide things in equations, you really must check that what you’re dividing by is not equal to zero. Okay let’s look at it a slightly different way. In line six, we’ve got 𝑝 minus 𝑞 times 𝑝 plus 𝑞, but 𝑝 minus 𝑞 is zero. So the left-hand side really means zero times 𝑝 plus 𝑞, and 𝑝 minus 𝑞 is zero on the right-hand side as well so that becomes 𝑞 times zero.

So we’re saying that zero times something is equal to something times zero. Well zero times something is zero, and something times zero is zero, so here we’re saying that zero is equal to zero. And yup, that is true. But that doesn’t mean that the things that we’re multiplying by zero are also equal. So moving from line six to seven, we don’t know that 𝑝 plus 𝑞 is equal to 𝑞. We only know that zero times 𝑝 plus 𝑞 is equal to zero times 𝑞. And that means that all of this is wrong; we haven’t proved that one is equal to two, we’ve proved that zero times one is equal to zero times two. Hooray! It looks like that the world of maths is safe after all.