Re-running the same contradiction on other prime square roots

0:00You have seen the argument that the square root of 2 is not a fraction. Now here is the square root of 3. And the square root of 5. It looks like a set of proofs to learn, one for each number. It is not. It is one proof. The number 2 never did anything in that argument except sit in a handful of lines and be prime. Take it out, leave a hole where it was, and you have a machine you can feed.

0:30This video builds that machine, feeds it, and then feeds it something that jams it. So go back through the argument and mark every place the number actually appears. You suppose the root equals a over b, with a and b sharing nothing. You multiply up and square, and you get: the number, times b squared, equals a squared. That is the first place it enters. The left side is that number times a whole number, so the number divides a squared.

1:04Then the step that needs a theorem: so the number divides a. You write a as the number times c, substitute, and cancel. And you land on: b squared equals something times c squared. Then the same theorem again, and the same conclusion about b. Every one of those lines carries the number and nothing else does. So rub it out everywhere and leave a slot. What is left is not a proof about 2. It is a proof about whatever you drop in.

1:37Drop in 3. Suppose the square root of 3 is a over b, sharing nothing. Multiply up and square: 3 b squared equals a squared. The left side is 3 times a whole number, so 3 divides a squared. 3 is prime, so 3 divides a. Same theorem, different prime. Write a as 3c. Then 3 b squared equals 9 c squared. Divide both sides by 3, and b squared equals 3 c squared.

2:10Look at that line. It is the equation we started with, with b in a's place. So 3 divides b squared, and 3 is prime, so 3 divides b. Both parts carry a 3, and they were supposed to share nothing. That is the contradiction, and not one line of it was invented for 3. Drop in 5, and this time say where the 5 goes rather than watching me write it.

2:38The equation after squaring is 5 b squared equals a squared. The theorem is used with 5, and gives that 5 divides a. Substituting a as 5c and cancelling leaves b squared equals 5 c squared. The theorem is used with 5 again, and gives that 5 divides b. Two uses, one prime, same collision. And notice what you were never asked. Nobody asked what the square root of 5 is.

3:10No decimal was computed. The argument does not know and does not care. So the statement worth carrying away is not about 2, or 3, or 5. It is this: if p is a prime, the square root of p is irrational. One statement, and every particular case is that statement with the slot filled. There are 46 primes below 200, and that is 46 results from one argument. But the statement says primes, and a careful person should ask why.

3:46Which step actually needs it? The usual answer is the theorem step, the one that says the number divides a squared, so it divides a. That answer is wrong, and the next two minutes are the most useful in this video. Test that step on 6, which is not prime. Does 6 dividing a squared force 6 to divide a? Take every whole number from 1 to 600 and check.

4:15It holds. All 600 times, with no exceptions at all. 6 is not prime, and the famous step does not care. The reason is that 6 is squarefree: no square bigger than 1 divides it. And squarefree is a much weaker condition than prime. From 2 to 200 there are 121 squarefree numbers, against 46 primes. So the theorem step is not what restricts the result to primes. Something else is.

4:49Run the argument on 6 and watch where it actually dies. 6 b squared equals a squared. 2 divides a, so a is 2c. Substitute: 6 b squared equals 4 c squared. Divide by 2. 3 b squared equals 2 c squared. And that is not the equation we started with. The shape is gone. The 6 has come apart into a 3 on one side and a 2 on the other, and there is nothing left to repeat.

5:21That is the step that needs a prime: not the theorem, but the repeat. Now the case worth seeing in full, because it fails in a different way. Drop in 4. 4 b squared equals a squared. The left side is even, so a squared is even, so 2 divides a. That step is fine. Write a as 2c. 4 b squared equals 4 c squared. Divide by 4. b squared equals c squared.

5:52Put that beside what 3 gave us: b squared equals 3 c squared. One line apart, and everything depends on it. The 3 version has no whole-number solutions at all. Search 40000 pairs and you find none. The 4 version has a solution for every single b. Take c equal to b, and you are done. There is no contradiction because there is nothing to contradict. And that is exactly right, because the square root of 4 is 2, which is 2 over 1.

6:29Now be very careful, because there is a wrong lesson sitting right there. The argument jammed on 4. It does not follow that the square root of a composite number is a fraction. It follows that this argument does not settle the question. Those are completely different things, and the second one is the damaging mistake. Look at what is actually true. Of the 153 composite numbers from 2 to 200, 140 have irrational square roots.

7:05Only 13 do not, and those 13 are exactly the perfect squares. 6, 8, 10, 12: all composite, all irrational. A tool falling silent is not the tool saying no. It is the tool saying: ask me something else. Which brings us to the second half, and to a question that looks harder and is easier. Show that 5 minus the square root of 3 is irrational. The instinct is to start again: suppose it is a over b, with a and b sharing nothing.

7:42Do that, and you will grind for a page and get nowhere. Because this question does not need a new contradiction. It needs one line of rearranging. Suppose 5 minus the root is a over b. Move the root to one side by itself. The square root of 3 equals 5 minus a over b. Put that over a common bottom: 5b minus a, all over b. Now look at what that is.

8:115b minus a is a whole number, because 5, b and a are whole numbers. b is a whole number, and b is not zero. So the square root of 3 would be one whole number over another. And we settled that ten lines ago. It is not. So the supposition fails, and no second argument was needed at all. Notice which property did the work there, because it is easy to miss.

8:40It was not anything about square roots. It was that whole numbers stay whole numbers when you subtract them. And that a whole number over a non-zero whole number is a fraction, by definition. That is what let the rearranged expression be recognised on sight. It is worth stating on its own: a fraction minus a fraction is a fraction. Checked across 9020 pairs, every single difference stayed one. So if a fraction plus an irrational number were a fraction, you could subtract the fraction back off.

9:18And the irrational number would be a fraction, which it is not. That one sentence is the whole content of the rule, and it is worth more than the rule. The same manoeuvre works with a multiplier, using division instead of subtraction. Show that 3 times the square root of 2 is irrational. Suppose it is a over b. Then the square root of 2 is a over 3b. Whole number over whole number, bottom not zero, and 3b is not zero because b is not.

9:53So the root would be a fraction, and it is not. Done. But this one has a condition the subtraction case did not, and it matters. The multiplier must not be zero. Zero times the square root of 2 is zero, and zero is a perfectly good fraction. So the rule is: multiplying an irrational number by a non-zero fraction leaves it irrational. Across 9020 combinations the reduction held every time the multiplier was non-zero, and the 220 with a zero multiplier are the cases the rule has to exclude.

10:35That is not fussiness. It is the one place the statement would be false. There are two general rules floating around at this point, and it is worth being honest about how much of them has been established. The first: a fraction combined with an irrational number by addition or subtraction gives an irrational number. The second: an irrational number multiplied or divided by a non-zero fraction gives an irrational number.

11:08Both are true. Neither has been proved here in general. What has been done is two particular cases, with the root already known to be irrational in each. That is not a criticism. It is what makes the two examples worth doing. If the general rules were already available, both would be one line of quotation. Working them out by hand is how you learn which property is being leaned on.

11:38So here are four to sort, and sorting them is the skill. 3 plus twice the square root of 5. One over the square root of 2. 7 times the square root of 5. 6 plus the square root of 2. None of them needs the contradiction machinery. Every one is a rearrangement. The first: suppose it is a fraction, subtract 3, divide by 2, and the square root of 5 is left alone as a fraction.

12:10The third: divide by 7. The fourth: subtract 6. The second one is the interesting one, and it is the only one that needs a moment's care. If one over the root is a over b, then the root is b over a. You have flipped it, and flipping needs a to be non-zero. Is it? Yes, and here is why: one over the square root of 2 is not zero, so a cannot be zero.

12:39That is the whole extra step, and it is the reason that item is set at all. Stand back and the shape is simple. At the bottom there is one argument, run once, that a prime's square root is not a fraction. Above it sit the particular primes, which are that argument with the slot filled. Above those sit results like 5 minus a root, or 3 times a root, which are not arguments at all.

13:09They are one rearrangement each, landing on a case already settled. So when you meet a question of this kind, the first thing to decide is which layer it lives on. Does it need the machinery, or does it need one line? Almost always it needs one line, and the line isolates the root. And if you ever want to check a root numerically first, you may, but know what it buys you.

13:361393 over 985 is the closest fraction to the square root of 2 with a bottom of 2000 or less. It is very close. It is not the root, and no fraction is. That is not what the proof rests on, and it never was.