Euclid's Elements, Book X, Proposition 18 (X.18)

Hello, this will be a demonstration of Proposition 18 from Book 10 of Euclid's Elements, which first of all says that if to the greater of two unequal straight lines, a rectangle equal to a fourth of the square on the lesser straight line be applied deficient by a square, and if it cut the greater into parts incommensurable with each other, then the square on the greater straight line is greater than the square on the lesser straight line by the square on a straight line incommensurable with the greater straight line. We're given a nearly identical setup to the previous proposition, Proposition 1017. In fact, what you'll notice about Proposition 1018, much like 1017, it really has two enunciations. It's the inverse and the contrapositive of Proposition 1017, so these two propositions are very much related to each other. 1017 about commensurability, 1018 about incommensurability. We'll have two unequal straight lines, BC and A, of which BC shall be the greater, and we will apply to the greater straight line, BC, the rectangle BD times DC, like so, such that it is equal to a fourth of A squared, and such that this defect right here, so the area by which the rectangle we've applied does not cover the whole straight line, but that area right there is a square. Furthermore, it'll be given that the part into which BC has been divided by the application, namely BD and DC, let these two straight lines, BD and DC, be incommensurable with each other. From this, we want to prove that BC squared is greater than A squared by the square on a straight line incommensurable with BC. Now, in just the same way as the previous proposition, namely if you take a look at my video on Proposition 1017, in exactly the same construction and demonstration in the first four steps of that proposition, we will be able to show that A squared plus DF squared equals BC squared. Remember how we define the straight line DF is that we bisected BC at E, and then on the other side of E, we make the straight line EF equal to DE, and that whole straight line there in the middle, FD, that is the straight line whose square is equal to the excess of BC squared over A squared. That is to say, A squared plus DF squared is equal to BC squared. Now, because BD was given as incommensurable with CD, we know from Proposition 1016 that the sum of those two straight lines is incommensurable with each of its parts. So because BD is incommensurable with CD, we know that BD plus CD, which is the whole line BC, we know that BC is incommensurable with CD. So that because CD is commensurable with BF plus CD, this was shown in the second step of Proposition 1017, that these two straight lines, BF and DC, they're equal to each other, so their sum is double CD as the ratio of a number to a number, 10-6, we know that they're commensurable with each other. So that because CD is commensurable with BF plus CD, and BC is incommensurable with CD, we know that BC is likewise incommensurable with BF plus CD. That's what Proposition 1013 tells us. If two magnitudes be commensurable, and either one of them is incommensurable with a third magnitude, then the other magnitude will be incommensurable with the same. So BC is incommensurable with BF plus CD. But because BC is incommensurable with BF plus CD, by Proposition 1016 again, BC must be incommensurable with BC minus BF minus CD. Now if we take BC, we subtract BF, and we subtract CD, that's just the straight line FD. So that because BC is incommensurable with BF plus CD by Proposition 1016, BC is incommensurable with FD. So that BC squared is greater than A squared by the square on a straight line incommensurable with BC, namely FD. And this is what we wanted to show. So we're done with the first part of this proposition. Now we'll move on to the second part of Proposition 1018. So now, on the other hand, conversely, if the square on the greater straight line be greater than the square on the lesser straight line by the square on a straight line incommensurable with the greater straight line, and if to the greater straight line a rectangle equal to a fourth of the square on the lesser straight line be applied deficient by a square, then the greater is cut into parts which are incommensurable with one another. So just as last time, we have two unequal straight lines, BC and A, of which BC is the greater. And to BC, we've applied the rectangle BD times DC equal to a fourth of A squared and deficient by a square figure, namely DC squared. Furthermore, let it be granted that BC squared is greater than A squared by the square on a straight line incommensurable with BC, from which we would like to prove that the parts into which BC has been cut, namely BD and DC, we would like to prove that BD and DC are incommensurable with one another. Now, just as was done in the first four steps of Proposition 1017, the previous proposition, we work to accomplish the same construction, defining DF in the way that E bisects BC, and then we cut off EF equal to ED, then the whole straight line FD will be such that A squared plus DF squared is equal to BC squared. Now, because we were given that BC squared is greater than A squared by the square on the straight line incommensurable with BC, and DF squared is the excess of BC squared over A squared, we can say from our given that BC is incommensurable with FD. That means, according to Proposition 1016, that BC must also be incommensurable with BC minus FD. That is to say, if we take BC, subtract FD, we're left with the sum of the straight lines BF and DC. By Proposition 1016, we know that because BC is incommensurable with FD, we know that BC is also incommensurable with BF plus CD. But, according to the second step of Proposition 1017, we know that CD is commensurable with BF plus CD, so that, according to Proposition 1013, BC is also incommensurable with CD, which then means, again, by Proposition 1016, that BC must also be in that, excuse me, BC minus CD must be incommensurable with CD. If the whole is incommensurable with one of its parts, then the parts are incommensurable with one another. That's what 1016 tells us, is that because BC is incommensurable with CD, it follows that the parts BC and DC, these are incommensurable with one another. BD is incommensurable with DC, which is what we wanted to prove. So, we're done with the second part of the proposition, and having demonstrated the second part of the proposition, we're therefore done with the whole proposition, therefore, etc., QED. Before launching into Proposition 1019, I'd like to say that the first 18 propositions, these are merely an appetizer for Book 10. We're about to get into rational straight lines, and the taxonomy of irrational straight lines, which is the real meat and potatoes, the real project of Book 10, is to classify irrational straight lines. We'll develop that project according to the propositions, and as we demonstrate those propositions, we'll be able to see more and more why this book was so important to Euclid, why it's an entire quarter of the elements. You read it, and it's seemingly quite dry, so it's hard to imagine what could have motivated Euclid to write such a book, and while that's ultimately impossible to say, we might be able to see into his intentions by just uncovering his project. So, before launching into those, I just want to say I'm really excited to get into those. There's a lot that I do want to share, and I hope that despite the dryness and the sort of drudgery of these propositions, it can really seem quite, what might you say, meaningless. At times, I really do promise that this book contains an immense amount of beauty, so I would just like to say that, because we're about to be on Book 10 for a while. With that said, that's all I've got to say about this proposition and the project ahead, so I'm going to end the video here and move on to Proposition 1019.