Two laws for adding, why they agree, and what addition obeys

0:00Someone walks from one point to a second, and then on to a third. Where have they got to? One move would have taken them there directly: the arrow from where they started to where they finished. That is the whole of the triangle law, and notice that nothing has been proved. It is a fact about walking. The word law arrives afterwards. Checked over eighteen walks, three starting points and six pairs of legs: the net move is the sum of the two legs every time.

0:34And it does not depend on where the walk began. The same six pairs of legs walked from three different starting points give six answers, not eighteen. The starting point cancels. Now the step that makes the law usable, and it is the only interesting step in it. The two arrows you want to add are almost never drawn nose to tail. They are just two arrows, somewhere on the page.

1:04So you pick one up and move it, until its tail sits on the other one's tip. Then you close the triangle. How often are they already arranged? Take six arrows drawn where they fell. Thirty ordered pairs among them. Not one of those thirty is already nose to tail. Every single one can be brought there by a slide. Thirty out of thirty. And the slide costs the arrow nothing: over the same thirty pairs the moved arrow's displacement is unchanged every time, and its tail is somewhere new every time. Same length, same direction, new place.

1:46That is the permission we were given earlier, and this is where it gets spent. A treatment that draws the after picture without the before has skipped the only part that needed saying. Next, something that looks like a second operation and is not. To take one vector away from another, you reverse the second one and add. That is it. There is no subtraction rule here, because none is needed.

2:16And the result gets called the difference, which is the right name, because a difference is a thing you end up with, not a separate move you make. So let us check that adding the reverse really does what a difference is supposed to do. A difference is the thing which, added to the second vector, gives back the first. So search the grid for that thing. No formula, just look for it.

2:43The search lands on the grid at one thousand two hundred and seventy two of the two thousand three hundred and four pairs, and refuses the rest because their answer falls off the grid it was given. And at every one of those one thousand two hundred and seventy two, it lands on exactly what reversing and adding built. Not once does it disagree. One operation. Two names for what you do with it.

3:13One warning about the difference before we move on, because it is the one thing here that is not symmetric. The two orders give different answers. Over those two thousand three hundred and four pairs, the two orders agree at forty eight of them. Forty eight. And the grid has forty eight nonzero arrows. So the pairs where the order does not matter are exactly the pairs of an arrow with itself, where both orders give nought. Everywhere else, reversing the order flips the answer.

3:48Hold on to that, because in about four minutes we are going to find that addition is not like this at all. Now the reason for a second law, and it is a good story. A boat is crossing a river. Its engine drives it straight for the far bank. The current carries it downstream. Both are happening at once. And that is the case the nose-to-tail picture does not obviously cover. Nothing here is a first leg followed by a second leg. There is no moment when the boat has done the engine and not yet done the river.

4:28Two vectors acting simultaneously, out of the same point, at the same time. The picture that fits is a different one, and it is the only place in this material where two vectors act together rather than one after the other. Draw both vectors out of one corner. Complete the parallelogram they span. The diagonal from that corner is the sum. That is the parallelogram law, and it is a claim about a specific segment: this diagonal, from this corner, not the other one.

5:06It suits the boat exactly. Both arrows start where the boat starts, both act at once, and the diagonal is where the boat actually goes. And now there are two laws for one operation, which ought to bother you. It should not bother you for long, because three lines settle it. Take the parallelogram. Walk the first side. Then walk the side that carries on from there. By the triangle law, those two add to the closing segment, which is the diagonal.

5:40And the side you walked second is the copy of the other original vector, sitting along the far edge of the figure. Substitute one for the other, and the triangle law has turned into the parallelogram law. Three lines. Checked over every pair of grid arrows that actually spans a parallelogram, from three different corners: the closing side and the diagonal are the same vector every time, and both are the plain sum of the two. No exceptions.

6:16But look at what that substitution rested on, because it is the third time today. The side you walked second is not the same segment as the vector it stands in for. They are in different places on the figure. They carry the same displacement — checked at every spanning pair. They never share a tail. They never lie along one line. They are equal vectors, and that is a definition that waives where each one starts.

6:47So: the triangle law needs an arrow slid. The parallelogram proof needs a side substituted. Both are cashing in the same idea, and neither argument mentions it. Now the question that the two laws make interesting. Not which law, but what addition obeys. Start with the order. Take the parallelogram again, and get to the far corner two different ways. Along the bottom then up the right side: that is the first vector plus the second.

7:23Up the left side then along the top: that is the second plus the first. Two routes, two different middle corners, one destination. So the two sums are equal, because they are the same segment. That is the parallelogram law read backwards, and it is worth saying so. Checked at all six hundred and twenty five pairs on a grid: addition gives the same answer both ways round every time.

7:53And here is why that is worth checking rather than assuming. Run the same test on subtraction, on the same twenty five arrows. It comes out the same both ways at twenty five pairs out of six hundred and twenty five, which is exactly the pairs of an arrow with itself. So the test is capable of saying no. It said yes to addition and no to the other one.

8:21Next, the grouping. Three vectors, laid along three legs of a path through four points. Add the first two, then add the third. Or add the first, then the last two. Both are the triangle law applied twice, and both land on the same closing segment. The only thing that changes is which triangle you look at on the way. Checked at all fifteen thousand six hundred and twenty five triples: the grouping makes no difference anywhere.

8:54And the control again: subtraction survives six hundred and twenty five of those triples. One in twenty five. So brackets around a sum of three are decoration. You can drop them, and that is not a convenience — it is a licence, and it had to be earned. Third. Is there a vector that leaves every vector alone? The zero vector, obviously — but let us count rather than assert.

9:25Exactly one arrow of the twenty five leaves every arrow unchanged when added from the left. Exactly one does from the right. And they are the same arrow. That word exactly is doing work. Run it on subtraction and you get a right identity — take away nothing and nothing happens — and no left identity at all. Which is why an identity has to be checked from both sides, and why the material states it in both orders on one line rather than one.

10:01Fourth, and this one was built two topics ago without being labelled. Every vector has an opposite: an arrow that sends it to the zero vector. Reverse it. And exactly one each. Every one of the twenty five arrows has exactly one partner that cancels it from both sides. The count of arrows with a different number of opposites is nought. Both sides, again. Is that fussy? Take a third operation, built to be lopsided: double the first arrow, then subtract the second.

10:38Under that one, twenty four of the twenty five arrows have no two-sided partner at all — while nine of them have a one-sided partner that looks perfectly good if you only check one side. Sixteen arrows where the two tests disagree. That is what checking both sides buys. So stand back and count what is on the board. Addition of vectors is commutative. It is associative. It has an identity, and that identity is unique. And every vector has exactly one opposite.

11:16There is a fourth thing, and it is the one nobody states: adding any two vectors gives you a vector. Closure. Worth measuring, because it is easy to think it means the answer stays on your grid. It does not. Of the twenty five arrows, adding pairs of them lands outside the grid two hundred and sixty four times. The closure is about arrows, not about the patch of paper you happened to draw.

11:46Those four properties together are the definition of one of the most useful structures in mathematics. All four are true at once, and they are almost never gathered up into a single list. You have just seen the list. One more result, and then the trap it sets. Walk all the way round a triangle: first side, second side, third side, back to where you started. You have gone nowhere. The three segments add to the zero vector.

12:19That is one substitution away from the triangle law — the closing side reversed is the negative of the closing side — and it is true at all fifteen thousand six hundred and twenty five triples of points we tested. Every single one. Which should make you suspicious, because not all of those are triangles. Here is the trap, and it is the most examinable false statement in this whole topic.

12:51If three segments add to the zero vector, are the three points a triangle? No. Of those fifteen thousand six hundred and twenty five triples, twelve thousand eight hundred and eighty eight are genuine triangles. The other two thousand seven hundred and thirty seven sum to nothing and are not triangles at all. And that is not a quibble about repeated points. Throw away every triple with a repeated point and nine hundred and twelve remain: three genuinely different points, summing to the zero vector, lying flat on one line.

13:31Here is one. Three points on a straight line. Walk from the first to the second, the second to the third, the third back to the first — and you are back where you started, because you walked out along a line and back along the same line. The equation holds. There is no triangle. So what does tell them apart? Compare the longest segment with the other two added. On the collinear one, the longest is exactly the other two together. On a real triangle, it falls strictly short.

14:07The equation is true in one direction only. Read it that way round. Now a drill. Four statements about one triangle. Three of them you can get from the triangle law by rearranging. Decide which one is not true. The first: the three sides in order add to the zero vector. The second: the first two sides added, minus the direct route, is nought. The third: the same thing again.

14:39The fourth: the first side, minus a reversed second side, plus the third. Take your time. Every one of them is true. Each was tested against all twelve thousand eight hundred and eighty eight triangles on the grid, and each is false at nought of them. So the question, as asked, has no answer. And this is worth knowing: sometimes the thing you cannot find is not there. The test can tell the difference, by the way. Change a single sign in the first statement and it fails at every one of the twelve thousand eight hundred and eighty eight.

15:20All four hold for one reason and no other: a segment read backwards is the negative of the segment read forwards. That one substitution turns any of them into any other. Last, the walker from the first scene, with numbers on their legs. Four kilometres due west. Then three kilometres on a bearing thirty degrees east of north. Where have they got to? The second leg cannot be added until it is split. Thirty degrees east of north is one and a half kilometres east and three root three over two kilometres north.

16:00The first leg is four kilometres west and nothing north. Add them. Two and a half kilometres west, three root three over two north. And the distance from where they started is the square root of thirteen kilometres. The squared length comes out at exactly thirteen, with no bracket left over. Which answers a question people get wrong constantly. The two legs were four kilometres and three kilometres. Four and three make seven. The displacement is root thirteen, which is about three point six.

16:39Not seven. Not even four. Root thirteen lies wholly below four, and they walked seven kilometres to get there. The length of a sum is not the sum of the lengths. One counterexample settles it, and we have one. But the grid can say something sharper. Over five hundred and seventy six pairs of arrows: the length of the sum never once exceeded the sum of the lengths. Not once.

17:08It fell strictly short at five hundred and thirty six of them, and it hit equality at forty. And those forty are exactly the pairs that point the same way. Not nearly — exactly: forty pairs give equality, forty pairs point the same way, and the two lists are the same list. So the answer is: only when the two point the same way, and never otherwise. That is a sharper statement than the one you were asked for, and it cost nothing but counting.

17:43Five things. The triangle law is a fact about walking, and the step that makes it usable is picking an arrow up and moving it. That move is the whole content of the free vector idea, and it gets spent on almost every figure here. There is one operation, not two. Subtraction is reversing and adding, and the difference is what you call the result. The two laws are one law. Three lines turn either into the other, and the substitution in the middle works because the two sides of a parallelogram are equal vectors — same length, same direction, different places.

18:27Addition is commutative, associative, has exactly one identity, and gives every vector exactly one opposite. Four properties, rarely met all at once, and you have now seen them together. And walking a triangle gives nought — but nought does not give you a triangle. Three points on a line do it too. The statement runs one way only. Everything in this topic except the last one is the same idea twice: it does not matter where you drew it.