What increasing and decreasing mean on an interval, before any calculus

0:00Here is a curve you have known for years. The square of the input, drawn against the input. And here are two columns of numbers taken off it. On the left, going from minus two towards zero: four, then two and a quarter, then one, then a quarter, then nothing. The heights are coming down. On the right, going from zero out to two: nothing, a quarter, one, two and a quarter, four.

0:28The heights are going up. You can see that. Anybody can see that. And seeing it is not the same as being able to prove it, which is the entire reason this topic exists. A picture settles the case for a curve you have already drawn. It settles nothing for a curve you have not. So what would a proof need? Not a picture, and not a table of five values either.

0:57It needs a statement about every pair of inputs at once. Something of this shape: take any two inputs from the interval, with the first below the second. Then compare what the rule does to them. That is the move, and it is worth pausing on, because everything after this point argues from inequalities and never from a graph again. The picture motivates. The inequality proves. And once you have the statement, there are five of them, not one.

1:31Here is the first. A rule is called increasing on an interval when, for any two inputs with the first below the second, the second output is at least as big as the first. At least as big. Look hard at that comparison, because there is a bar under it. Not greater than. Greater than or equal to. And the second part is the mirror image. A rule is called decreasing when, for any two inputs with the first below the second, the second output is at most as big as the first.

2:08Again a bar under the sign. That bar is one stroke of ink and it is the whole of this topic. It is the difference between a flat stretch counting and not counting. So let us see what it lets in. Take the simplest rule there is. The one that gives the same answer whatever you feed it. A horizontal line. Pick any two inputs with the first below the second.

2:36The two outputs are equal. Now ask the first question: is the second output at least as big as the first? Equal counts as at least as big, so yes. The rule is increasing. Now ask the second: is the second output at most as big as the first? Equal counts as at most as big too, so yes again. The rule is decreasing. A rule that never moves is increasing and decreasing at the same time, and I did not choose to say that.

3:11The definition says it, and running the definition over twenty-four different situations says it back. Those two, the horizontal line on a closed stretch and on an open one, are the only two of the twenty-four that come out both ways. There is a third part, and it is the easy one. A rule is constant on an interval when it takes one value throughout. Which our horizontal line also is, so it is now three things at once.

3:44And then there are two more parts, and they are where the bar comes off. Strictly increasing: for any two inputs with the first below the second, the second output is greater than the first. Strictly decreasing: less than. Set them side by side and one character moves. Increasing allows equal outputs. Strictly increasing forbids them. Which is why the horizontal line, which passes both of the first two, fails both of the last two.

4:17It is increasing, decreasing and constant, and it is neither strictly increasing nor strictly decreasing. Now here is the thing worth staying awake for. When you are shown what neither increasing nor decreasing looks like, the picture you will usually be given is a flat segment with a solid dot at each end. A horizontal line on a closed interval. That is exactly the object we just tested. And we tested it against the definitions.

4:51It came out increasing and decreasing. Not neither. Both. The caption on that picture is right about the two strict notions and wrong about the two it actually names. I am not asking you to take my word for it. Run the statement. Take two inputs from the segment, compare the outputs, and ask whether the second is at least the first. Every single time, it is. You are not going to be examined on that collision.

5:24But if you never notice it, then one day you will be asked why a constant function is classified one way in the words and the opposite way in the picture, and you will have nothing to say. The two families are not independent. If the second output is always strictly greater, then it is certainly always at least as great. So strictly increasing implies increasing, and strictly decreasing implies decreasing.

5:55Across all twenty-four situations, not one is strictly increasing without also being increasing. Not one. Going back the other way is a different story. It fails at five of the twenty-four, and those five are exactly the ones with a flat stretch somewhere in them. This matters for how you write an answer. If you are asked to show something increasing and your working ends up establishing the strict version, you have not made a mistake.

6:27You have proved something stronger than you were asked for, and the weaker claim follows. But the two words are not interchangeable, and one of them will be asked for by name. Here is a proof, done properly, with no calculus in it. Show that seven times the input, less three, is increasing everywhere. Take two inputs, call them a and b, with a below b, and do not give them values.

6:56That is the whole trick. Multiply both by seven: seven a is below seven b, because seven is positive. Subtract three from both: the order is untouched. So the output at a is strictly below the output at b. That is every pair, in three lines. And notice what it did not need: not a derivative, not a graph, not a single number. The slope alone settles it, and the same argument runs for three times the input plus seventeen, and, with the sign turned round, for one less twice the input, which is strictly decreasing.

7:39Now for the thing students actually do, which is to try a few numbers. Let us be precise about what that can and cannot buy you. Suppose you check some pairs of inputs and one of them breaks the condition. Then the claim is dead, and you have a proof that it is dead. Trying inputs refutes. Now suppose you check some pairs and they all hold. You have learned that those pairs hold.

8:06You have learned nothing whatsoever about the pairs you did not check. Trying inputs does not establish. And even refuting takes a little care. The square on the stretch from minus two to two takes the value four at both ends, so if the only two inputs you try are the two ends, the outputs are equal, equal is allowed by the first part, and you conclude it is increasing.

8:34Add one input in the middle and it collapses at once. I want to show you how badly this can go, with a number attached. Take the square of the input less one thousandth, on the stretch from zero to one. It turns one thousandth of the way in. For that first thousandth it is falling, and after that it rises for the rest of the stretch. So it is neither increasing nor decreasing there.

9:04Now put a ladder of twenty-five evenly spaced inputs across it and run the definition. The rungs are about a twenty-sixth apart. Not one pair of them lands inside that first thousandth. Every pair rises. The ladder reports, confidently, that the rule is increasing, and strictly increasing too. It is wrong about both. I put that same rule to eight different ladders, from five rungs up to five hundred. Seven of the eight get it wrong.

9:38It takes five hundred rungs before two of them land close enough together to see the descent at all. So how do I know the ladder is wrong, if a ladder is all I have? I do not use a ladder. For any rule that is a square or simpler, there is an exact way through, and it is rather pretty. Take two inputs a and b with a below b.

10:03The difference between the outputs factors: it is the gap between the inputs, times a straight-line expression in their sum. The gap is positive, always, because a is below b. So the sign of every single difference is the sign of that straight-line expression, evaluated at the sum of the two inputs. And the sum of two inputs from an interval ranges over an interval of its own, running from twice the low end to twice the high end.

10:36That is it. A statement about every pair, of which there are infinitely many, has become a statement about the range of one straight line, which is two subtractions. Twelve of the twenty-four situations are simple enough to go through that door. Twelve situations, five parts each, sixty verdicts, and the two doors agree on fifty-eight of them. The two they disagree about are both at the square that hides its turn a thousandth in, and it is the exact door that is right.

11:13That should not leave you thinking a ladder is useless. Move the turn from a thousandth of the way in to a twentieth of the way in, and everything changes. The same twenty-five rungs now straddle the descent, the definition is refused, and the ladder agrees with the exact answer. The only thing separating the two cases is whether two rungs can land inside the falling part. That is the honest summary of what checking values gets you.

11:46It is not that trying inputs is cheating. It is that whether it works depends on something you do not know until afterwards, which is exactly where the interesting behaviour is hiding. Now the word neither, properly. Take the cosine over half a turn. Every pair of inputs you can find has the second output below the first, so it is strictly decreasing there. Over the next half turn, every pair rises, so it is strictly increasing.

12:19Now put those two stretches together and take the whole turn. There is a pair of inputs whose outputs fall. There is another pair whose outputs rise. The first pair kills every claim about rising. The second kills every claim about falling. So on the whole turn, all five parts are refused, and that is what neither means. It does not mean the rule does nothing. It means both things happen inside one interval, so no single verdict can cover it.

12:52The sine says the same over its two quarter turns and over the half turn that contains them both, and so does the square about zero, and so does the size of the input, whose turn is a sharp corner rather than a smooth one, which makes no difference at all to the verdict. There is one more definition and it is shaped differently. Rising at a point. And read it carefully, because it is not a statement about the point.

13:23A rule is said to be increasing at a point when it is increasing on some open interval containing that point. Some interval. You get to choose it, and you only need one to work. So the object of the claim is a neighbourhood, and the point is only the thing the neighbourhood is around. Why define it that way? Because at a single point there is nothing to compare.

13:50Every one of these definitions needs two inputs, and one point does not give you two. Here is why the distinction earns its keep. Consider a rule that agrees with the input when you are far from zero, and turns back on itself close to it. Feed it minus one and you get minus one. Feed it zero and you get zero. Feed it one and you get one. Read off those three and it looks like the graph is climbing straight up through the origin.

14:24It is not increasing at zero. I offered it six intervals around zero, from a whole unit wide down to a ten-thousandth, and it fails on every one of them, because each of them contains a pair of inputs whose outputs go the wrong way. Ask the same question at a point away from the turn and the answer is yes at once. Going up through a point and being increasing at a point are two different claims, and only one of them is the definition.

14:58Now something practical, and slightly awkward. The short summary version of these definitions, the one you are most likely to revise from, is not the same as the definitions themselves. It gives increasing and decreasing with strict comparisons. No bar under the sign. So it is quietly handing you the last two parts under the names of the first two. Run both versions over the same twenty-four situations and they part company at five of them.

15:31Those five are the same five with a flat stretch in them, which is not a coincidence. It is the only place the two versions can possibly differ, because the bar is the only thing that changed. And there is worse, because that same short version disagrees with itself. Alongside the strict comparison it offers an alternative test for increasing: that the rate of change is at least zero throughout the interval.

16:03At least zero. Which permits a rate of exactly zero, which permits a flat stretch, which the strict comparison one line above forbids. Those two cannot both be the definition of the same word. Here is one rule that separates them. Flat to the left of zero, and rising as the cube to the right. I measured its rate of change at eleven inputs across the stretch. Every one settled, every one is at least zero, and at the six inputs from the flat side it is exactly zero.

16:40So the derivative test says increasing. And the strict comparison, run over the pairs, says no, because two inputs from the flat part give equal outputs. One rule, one interval, two answers, from two halves of the same sentence. So what do you actually write down? Quote the five-part version, with the bars where they belong, and know which of the five you are being asked about. If the question says increasing, the bar is in.

17:13If it says strictly, the bar is out. Prove the strict version when you can, because it gives you the other one for free. Never argue from a picture, and never argue from a handful of values, because a handful of values got the square that hides its turn wrong seven times out of eight. Take two inputs, do not give them numbers, and compare the outputs. That is the whole method, and it is the same three lines every time.

17:43Three things to carry away. The bar under the sign is the topic: it is what lets a flat stretch count, and it is what makes a rule that never moves increasing and decreasing at once, in two of the twenty-four situations here and in no others. Strict implies not-strict and never the reverse, which failed at five of the twenty-four. And a statement about every pair is not checked by trying pairs; it is checked by an argument that covers all of them at once, which is why the exact route settled fifty-eight of sixty verdicts without ever looking at a single pair.