 The elements of Euclid consist of 13 books, we'd call them chapters, and hundreds of propositions. We'll go through all of them. Eventually. But for now, we'll focus on a few of the important ones. Now, in the 5th century, Proclus wrote a commentary on Euclid's elements, and he classified the propositions in the elements as one of two types. He either called them problems, which we might view as techniques for doing a certain geometric task, and theorems, which are more abstract results. So let's take a look at the first three propositions in Euclid's elements. The very first proposition of the elements is to construct an equilateral triangle. And here's Euclid's solution. Given our line segment AB, we'll draw a circle with center A and radius AB. Likewise, we'll draw a circle with center B and radius AB. The intersected C and ABC forms an equilateral triangle. So this is a construction. What makes it a proposition is two things. First of all, Euclid proves that you could do every single one of these steps. Let's take that apart. So we have our line segment. Draw a circle with center A and radius AB. Well, that's the third axiom. About any point, and with any given length, a circle may be drawn with the point as center and the length as radius. So these circles exist. What about the size of the triangle? Well, the very first axiom tells us that we can join any two points with a straight line. And that means this triangle ABC exists. So lastly, we need to show that we have an equilateral triangle. So AC is equal to AB since both are radii of the same circle. BC equals AB since both are radii of the same circle. And our common notion says that things that are equal to the same thing are equal to each other. So AC is equal to BC. Now, because this proposition resulted in the construction of an equilateral triangle, this is a problem in Proclus' categorizations. However, because of the way that Euclid proved it, it can be restated as a theorem. In this particular case, given any length, an equilateral triangle exists with sides equal to the given length. And this is generally true. Every proposition in Euclid that corresponds to a geometric construction can be reframed as a proposition that claims the existence of a certain type of geometric figure. Let's look at Euclid's second proposition. And this goes to a standard technique in Euclidean geometry, which is to use a compass to move a length. But can we really do that? And this actually requires a proposition to construct a line at a given point equal to another given line. And here's Euclid's construction. Suppose we want to construct at A a line equal to BC. So first, we'll construct an equilateral triangle, ABD. And we know we can do that because of the first proposition. Next, we'll construct a circle with center B and radius BC. And again, we know we can do that because of our axioms. We'll extend DB to G. And we can do that because this video is on YouTube. Wait, no, actually that's one of our axioms. Our second axiom says that we can extend a straight line in a straight line. We'll construct a circle with center D and radius DG. That's our third axiom. We'll extend DA to L. Again, that's extending a straight line in a straight line, our second axiom. And Euclid claims that AL is the required line. And the proof goes as follows. DL is equal to DG because they're both radii of the same circle. DB is equal to DA because they are two sides of an equilateral triangle. And AL is equal to BG. And this comes from our common notion. DL is equal to DG and DB is equal to DA. So when we remove equals from equals, the results are equal. Once again, BG is equal to BC because they're radii of the same circle. And things equal to the same thing are equal to each other. And so AL has the length BC. And again, even though this is a problem, we can restate this as a theorem. A line segment can be moved to begin at another point. The third proposition of the elements is to cut off a line of a given length from a given line. So as a problem, given line AB, cut off a line equal to C. So first, we'll make AD equal to C. We can do that because of proposition 2. Construct a circle with center A and radius AD. That's axiom 3. And the claim is that EB is the required line. And that's because AD is equal to AE because they're both radii of the same circle. AD is equal to C. Things equal to the same thing are equal to each other. So AE is equal to C. And so EB is the line AB with the line C removed.