### Video Transcript

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

First of all letβs define two variables, π and π. And weβll say that the value of π is equal to the value of π. Now that gives us the equation π is equal to π. Now letβs take that equation and multiply both sides by π. So π times π is equal to π times π or simply ππ is equal to ππ. But wait! ππ just means π times π or π squared, so we can say π squared is equal to π times π or ππ.

And now letβs subtract π squared from each side of that equation. So weβve done the same thing to both sides of that equation, so itβs still an equation. And π squared minus π squared is equal to ππ minus π squared. Now when weβve got a square thing minus another square thing, we call this the difference of two squares. And we can factorize that expression like this: π squared minus π squared is equal to π minus π times π plus π.

So letβs just check that: positive π times positive π is positive π squared; positive π times positive π is positive ππ; negative π times positive π is negative ππ; and negative π times positive π is negative π squared. Now ππ means π times π, and ππ means π times π. But multiplication is commutative, and that means that it doesnβt matter in which order you multiply things together in. So π times π gives you the same result as π times π, and that means that I can re-express ππ as ππ.

So our expression now reads π squared plus ππ minus ππ minus π squared. And ππ takeaway ππ is nothing. And in multiplying out π minus π times π plus π, Iβve 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 Iβve now got π minus π times π plus π is equal to ππ minus π squared.

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

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

Now remember, back at the beginning we said that π was equal to π, and that means I can replace π with π in my equation, which gives us π plus π is equal to π. Well π plus π is just equal to two π. So now two π is equal to π. Now I can Divide both sides by π; and on the left, π divided by π is one; and on the right, likewise, π divided by π 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; weβve 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 isnβt equal to two. If it was, I think weβd have heard it on the news by now. So letβs go through our working out line by line and see if we can work out whatβs happened. Well defining two variables π and π and letting them be equal, thereβs no problem with that. And itβs perfectly okay to multiply both sides of our equation by the same thing π. And yes, π squared is equal to π times π, so thatβs correct.

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

Now in moving from step six to step seven, we divided both sides of the equation by π minus π. Well this might normally be okay, but we did say at the beginning that π was equal to π. So π minus π is equal to something minus itself; thatβs zero. So we are dividing both sides of our equation by zero. Now that is a problem. For example, whatβs one divided by zero? Well it doesnβt matter how many times you add zero to itself, youβre never going to reach one.

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

So weβre 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 weβre saying that zero is equal to zero. And yup, that is true. But that doesnβt mean that the things that weβre multiplying by zero are also equal. So moving from line six to seven, we donβt know that π plus π is equal to π. We only know that zero times π plus π is equal to zero times π. And that means that all of this is wrong; we havenβt proved that one is equal to two, weβve proved that zero times one is equal to zero times two. Hooray! It looks like that the world of maths is safe after all.