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This video explains how to prove a mathematical proposition. It shows how two parallel lines, a straight line divided into equal and unequal pieces, and a square can be used to show that the rectangle formed by the unequal pieces plus the square of the difference between the equal and unequal pieces is equal to the square of half the line. Algebraically and intuitively, it is demonstrated that the product of the two line segments gets smaller as the distance between them increases. Watch to learn the full proof!
A straight line AB is cut into equal pieces AC and CB, and also into unequal pieces AD and DB. The proposition states that the rectangle formed by the unequal pieces, plus the square of the difference between the equal and unequal pieces, is equal to the square of half the line AB. This is proven by drawing a square on one of the pieces, then drawing parallel lines through D, H and A. The area of CM is equal to the area of DF, which is equal to the area of the square DB added to CH and CF respectively. Therefore, the green area plus the black area form the red area, as per the proposition.
Two parallel lines AP and KM have been constructed, and AC is equal to CP. This means that the areas of CM, AL and DF are all equal. The rectangle AH and the Neumann have equal areas, which can be shown by decomposing them into two equal parts. A line is divided into two unequal pieces, which can be shown algebraically and intuitively. It is shown that the rectangle contained by the unequal pieces plus the square on the difference is equal to the square on the half the length, meaning the product of the two line segments gets smaller as the distance between them increases.
A straight line AB is cut into equal pieces AC and CB, and also into unequal pieces AD and DB. The proposition states that the rectangle formed by the unequal pieces, plus the square of the difference between the equal and unequal pieces, is equal to the square of half the line AB. This is illustrated by the diagram, with the square of CD being equal to the square of half the line.
A line is cut into two pieces and a square is drawn on one of the pieces using Proposition 1.46 from Chapter 1 Book 1. Parallel lines are then drawn through D parallel to the square, H parallel to AB, and A parallel to BF. The intersection points of these lines are labeled and Proposition 1.31 is used to construct them. Finally, the lines are marked as parallel.
A rectangle ADCB and a small square DB are drawn. The claim is that when the green area is added to the black area, the red area is formed. This is proven using Proposition 1.43, which states that the area of the complements of a parallelogram along a diagonal are equal. Proposition 1.36 is then applied, which states that parallelograms between the same parallel lines with equal bases have equal areas. Therefore, the area of CM is equal to the area of DF, as they are both equal to the areas of the square DB added to CH and CF respectively.
Two parallel lines AP and KM have been constructed, and AC is equal to CP. This means that the areas of CM, AL and DF are all equal. The rectangle AH can be decomposed into AL and CH, and the Neumann can be decomposed into DF and CH. Both the rectangle and Neumann have equal areas.
A rectangle with sides a and h, and a square with side CF, have equal area. This can be shown by decomposing the rectangle into two equal parts, Al and DF, and the square into the Neumann CFmoreLG and LG. Since h is equal to DB, a h is equal to 80 cross DB and LG is equal to the square on CD. The Neumann CFmoreLG plus LG is equal to CF, so a h is equal to CF. This proves Euclid's 2.5, that the rectangle and the square have equal area.
A line is divided into two unequal pieces, labeled A and B, and then equally, labeled C and D. Algebraically, this proposition states that the rectangle contained by the unequal pieces plus the square on the difference is equal to the square on the half the length. Intuitively, this means that the product of the two line segments gets smaller as the distance between C and D increases, and the product is at its maximal value when the two things being multiplied are equal. This can be seen as a constraint on the function Z times X Y, with Y equal to a fixed value Lambda minus X.
cool okay so this is where the um the Neumann appears I'm not even sure if this is the right way to pronounce it but I'll say I think I'll stick with Neumann If it's incorrect I'm sure we'll um we'll get lots of YouTube comments of people telling me I'm wrong okay so proposition 2.5 so this one um is a little bit longer state so um I'll draw the picture after I um I State the proposition so if a straight line is cut into equal and unequal pieces I have to confess this um this bit of the sentence caused me a little bit of confusion the first time I read it uh is um only until I drew the picture that I realized oh I looked at the picture didn't draw it myself look at the picture um did I realized what was meant so I'll clarify this as we go actually maybe I'll clarify this now because um uh I'll draw the picture under here so straight line AP so a b is cut into equal pieces so let's just focus on the first thing first part of the sentence first so we're going to cut it into equal pieces meaning that AC is equal to C B so uh this is what they mean by cutting a b straight line into equal pieces so I'm just going to put the Strokes here to indicate that AC well maybe I'll just write down AC is equal to C B okay so this is the equal pieces and then it's also cut into unequal pieces so we're going to have a point D different from C it doesn't matter which side is on and uh here AC sorry a d is not equal to DB but this is the unequal pieces okay so I cut the line like this so we get a bunch of line segments lying on a b then the rectangle is obtained by the unequal pieces then so now we can state uh hypotheses and conclusion so then the rectangle hopefully I can fit the proposition with the space I've left myself contained by the unequal pieces okay so now that we've got um the diagram I'm going to call this a d cross DB I'm just going to indicate this this is not notation used for Euclid but you all know what I mean when I write a d cross t b uh DB Plus the square on the the square on the difference between the equal and unequal pieces so I'm just going to mark this so this is the bit we're talking about um plus the square on I mean I'm really tempted to just write CD here but I'll write the whole thing out plus the square on the difference between the equal and unequal pieces so this is CD Cross C D this is the square on CD is equal to the square and half the line Okay so we've got two things right one rectangle uh one small square and then we're going to say it's equal to a big
square and it's equal to the square on half the line foreign but maybe it's worth remembering that there's a proposition in book one that says we can draw given a point on a line and another line a segment on the first line of length equal to the length of the second line so I can always transplant a line into some other line so that the difference between two line segments is a well-defined thing by that kind of process do you remember the number no no that's what the bot is supposed to be for but I don't think it's good enough oh is that right well that'll be great if the boy can just tell us I certainly don't remember the number it's uh okay I think that's a good point right because when you say difference we are doing something of the nature of what you said cool thanks Dan okay so um so the construction is kind of uh half complete we'll complete the rest of it so now um we've got you've got the line we've cut into pieces and what we're going to do the first thing we're going to do is we're going to use our trustee squared proposition 46 from chapter one book one to draw a square on CP so uh CB is uh is here so uh C be I'm going to make a square here and here this is a proposition 1.46 right Square on CB okay so we were a bunch more constructions to go we have to join up this line This diagonal and we're going to label some stuff this Square CB we're going to call c e f b [Music] I can still write the cefb and um the next thing we're going to do is we're going to draw a bunch of parallel lines we're going to draw parallel lines um through d parallel to BF so um remember there was a proposition 31 that allows us to draw all kinds of parallel lines through points a parallel parallel to existing lines so I'll write that in in a bit so there's this one we're also going to draw a parallel line from we're going to draw a line actually through H so this is H here is in a section of d g and I G is the point where the line through D parallel to BF or c e hits e f so this Mark these in and we're going to draw another parallel line through h parallel to a b c so I'm going to try and just do this this eyeball it here there we go right and then we're going to draw a final parallel lines through a parallel to b f and then where the label these intersection points uh like so I think I got all of them and let me just Mark these two as parallel and then this one is parallel so proposition 1.31 that allows us through all these right construct parallel lines through uh we did D
h and a I guess parallel to the things that uh does indicated on the diagram so I won't write all that out in a small space that we've got okay so this is the uh the construction that we need and I'm just gonna do one more thing um just to make it uh kind of clear what we're doing so this is not in in Euclid I'm just going to draw it in for uh for our sake so what's the uh statement here the statement here is that um the rectangle a d cross D B so a d uh well a d cross DB is going to be this thing here or maybe so this thing here is a d cross DB and the big uh the big Square is I guess c b a c or CP where is this guy here right and the claim is that um the climate order and then there's also this this uh this small square down uh I'm running out of colors down here this is the small square here right the claim is that when you take the green um and you add it to the black you get the red right so um it may help to just kind of indicate that you can subdivide this rectangle here with this line and uh right you can kind of play some jigsaw puzzles in your mind to convince yourself that the statement is true but of course there's not a proof so now we'll go off and improve this so the proof goes as follows okay so by proposition 1.43 the area of ch is equal to the area of h f so CH is um so these are the the uh the complements of um of a parallelogram so CH going back to the diagram here I think I'm shaded in a few too many things right but CH is this guy here right and um HF is this guy here right so these are complements uh of a uh parallelogram along a diagonal and so this is exactly what proposition 1.43 gives us that the area of CH is going to be equal to uh to HF okay so so uh therefore the area of c m is equal to the area of DF right so let's go back why is that well c m right is this thing here right CH plus the square DB and uh DF is this guy here and this is HF plus the square DB so right you're adding the same thing to two equal things so therefore they must be equal so yeah since they're equal to the areas of the square DM added to c h and C F respectively so I think I'll leave that explanation now I can I guess I just said it so um by proposition 1.36 so I'm going to record this proposition right this proposition says to the parallelogram between the same the same parallel lines with equal bases have equal areas thanks for the lines with equal basis have equal areas okay so I guess uh how we're going to apply this the area of cm is equal to the area
of a l okay so going back to the diagram cm is this one here and a l is this one here right so why are the equal right the between parallel lines uh that between the parallel lines AP and km these two lines are constructed to be parallel and by construction AC is equal to CP right so these are parallelograms or actually rectangles with equal bases for the basis being AC and CB between two parallel lines so they have equal areas and so um so then we can conclude um we can conclude that the areas of c m a l and DF are all equal right so uh yes cm DF and a l right they're all you know transitivity property of equality there right so they're all equal so what we're going to do with that um next the rectangle a h a h can be decomposed into Al and ch okay going back to the diagram um the I guess this is uh uh different color so um the rectangle ah is this big one here right all we're saying is that we can write this as a l um and ch right because they're kind of next to each other put them together you get a big rectangle okay it can be the composition to it and the Norman so the Euclid has a funny notation for the Norman I'm gonna use um slightly more modern rotation to say what it is so the Norman is you take a square and you kind of knock out the uh knock out the square on the opposite on one of the diagonals right so if you look at this Square here c c f right if you take CF and you cut out this black bit down here right so then the the remaining with the complement of the big Square by the little square is annoyman and equivalently you can well not equivalently analogously you can remove the big Square DM and get another annoyman which is like here right so because that's what we're actually doing I'm just going to call it what it is CF complement LG right uh in the moment CF um sorry I'm just looking at my noise it I seem to have missed hang on let me let me say that so the rectangle ah symbicating imposed into a l and C H right so you've got a l and CH and then you've got a Neumann here um on the other side all right sorry so the ah can be decomposed into a l and CH and the Neumann uh can also be decomposed uh into DF and ch right so why is that so you go back to here the Neumann we're talking about is the uh the big squared minus the little square um and so we can decompose it as DF and CH and of course this is just because DF is here and CH is here all right so we're just cutting it up into two rectangles okay so now uh hence they have equal areas okay so what do I mean uh so the Norman
here has equal area with just pH right so these are the ones which have equal areas because we can decompose them into uh ch right this is the same and Al and DF and we kind of set the Al and DF are the same so then we've basically cut these um these geometrical figures up into equal things that we've shown to be equal and so this is uh so the main conclusion here is that the rectangle ah has equal area with the Norman cfmore CF complement LG so um I think we're almost there um so since the h is equal to DB um a h is equal to well equal to 80 cross DB not only equal to right it's um it's I guess not only did it have equal areas as you know a h um they have equal sides right one on the sides of AHS is um equal to 80 and the other one d h is equal to DB right so there's that also LG is equal to the square on CD right so LG is a little black square that I shaded black on the main diagram uh I know I probably should walk across here but then I can't see the diagram anyway so LG is um the little black square that we cut out of the Norman and this is the square and CD remember it's the square on CD was one of the um uh uh things that we had to in well that was one of one of the um things in the statement so finally the Norman CF oh that's not what I said CF more LG plus LG is CF I guess this is um kind of a tautology if I write it like this right I'm just going to put the LG Square back in you get the whole Square back so hence uh ah is equal to a d cross DB which is what we said above Plus LG um is equal to CF mod LG plus LG which is equal to CF okay so um let me just say yes so this this is the um yeah so this is this thing here is a square on CD right so remember the statement was that we take the rectangle a h we add the little square on which is the square on CD which we see is equal to LG and we're going to get the uh the square right and we'll decompose a square into the Neumann plus LG and we've kind of shown before that the Neumann and ah the same we add LG on both onto both of them and we get the statement of of Euclid 2.5 okay so uh that was actually longer than I thought right we took up the most of the half hour yeah it's fun I think uh maybe I can follow on with that making some remarks on its relation to the algebraic form um and sort of yeah thinking about it while you're speaking I think I have a bit of a better idea how to intuitively think about this somewhere between the algebra and the geometry you want to give me the speaker for a minute yeah
okay um all right so let me redraw the diagram and and label some things and then just comment on the algebraic form so if you divide a line into two unequal pieces I'm going to call one of them a and one of them B then if you were to divide it equally I guess I'll put it there so that was c and d right so that's a that's B that's C and that's d Okay so algebraically the proposition says a b the rectangle contained by the unequal pieces of the hole plus the square on the difference well the difference is that which is the difference between a half a plus b and well either one um so the smaller one the way I've drawn it is is B I guess that's actually I mean the picture sort of has to be on that side right uh yeah so that's maybe a worth observing um is equal to the square on the the half the length so a half a plus b squared uh I'm sorry rewriting that that says that it's a plus b on 2 squared minus a minus B on 2 squared or if you want to make a substitution X is a plus b on 2 Y is a minus B on 2 then what this says is x squared minus y squared is x minus y plus plus y oh this is much nicer than what Leroy yeah so but there's an interesting kind of there's more than this expression contained in this observation because if you look at think about it in the following Dynamic way so imagine taking a line splitting it in half at C and then considering the square on that half right so that's that's our starting point so that's here and then you can imagine sliding that point C I mean you sort of you're sort of imagining these two line segments right like the the left and right hand segments of wherever you cut that line a b and then you move that cut C over to D so now you're taking the product of those two bits and so you've got the square rectangle rather once you shift that right so now you've got a D times DB and you'd like some intuition for for example does that get bigger or smaller as you take D away from C now the the kind of even more than the formula the intuition that you get from from this equality here is that indeed it gets smaller right right and I suppose the further information is how much smaller right it says that a b is actually less than a half a plus b squared so the square is that it's well the product is at its maximal value and the two things being multiplied are equal and that I mean we would recognize today that as being something about the function Z times x y and you have the constraint that Y is equal to some fixed fixed thing Lambda minus X and you would