11. Let T be the statement: The sum of any two rational...

11. Let T be the statement: The sum of any two rational numbers is rational. Then T is true, but the following "proof" is incorrect. Find the mistake. "Proof by contradiction: Suppose not. That is, suppose that the sum of any two rational numbers is not rational. This means that no matter what two rational numbers are chosen their sum is not rational. Now both 1 and 3 are rational because 1 = 1/1 and 3 = 3/1, and so both are ratios of integers with a nonzero denominator. Hence, by a supposition, the sum of 1 and 3, which is 4, is not rational. But 4 is rational because 4 = 4/1, which is a ratio of integers with a nonzero denominator. Hence 4 is both rational and not rational, which is a contradiction. This contradiction shows that the supposition is false, and hence statement T is true." 12. Let R be the statement: The square root of any irrational number is irrational. a. Write a negation for R. b. Prove R by contradiction. 13. Let S be the statement: The product of any irrational number and any nonzero rational number is irrational. a. Write a negation for S. b. Prove S by contradiction.

Transcript

00:02 Okay, let's start with this first one.

00:04 So the first one is a classic mistake between inductive and deductive reasoning.

00:10 There's a difference between inductive and deductive.

00:14 Right, so what we're doing here, whenever we do a mathematical proof, we're using deductive reasoning to show that all cases are true.

00:22 So in the example of this proof, it says that now we have the numbers one and three, right? we have one is one over one, three, three over one.

00:31 So in this proof, they're trying to use inductive reasoning by showing that there exists some x and a y, right? but that immediately is the wrong format, the wrong approach to making a proof because what we're trying to do is use deductive reasoning to show that for every x and every y.

00:52 Right? just to show that there are two numbers that for this to be true is not enough.

00:58 And then moreover, they say that the ratio, the sum of one and three is not rational.

01:06 It's four.

01:07 It is.

01:07 It is rational.

01:08 So you can't just suppose that it is not rational.

01:12 You have to show that it is not rational.

01:15 But it is by definition rational.

01:17 Four is a rational number.

01:18 So that's the main problem with this proof is that it just shows two examples of one and three.

01:28 There's some being four, and then says that, but four is not rational because we suppose that it wasn't.

01:34 That's not the way that we're doing it.

01:36 What we have to do is we have to show that for every x and y, this is the truth.

01:42 And we have to do it by contradiction.

01:44 We have to suppose not and say that we have to say that suppose that we have these two rational numbers, x is some a over b where a and b are integers, and y is a, c over d where c and d are integers, but a or x plus y is not rational, right? that's how we have to approach it.

02:10 So then we would go by and we'll say, well, when we show it, it actually turns out that it is rational, but we assumed that it wasn't.

02:17 And that would be the method to do it.

02:19 We have to use definitions to show that for every x and y, which are defined as rational numbers.

02:29 Then and then show that when we add them together, we get another rational number, but we assumed that it wasn't.

02:36 That would be the case.

02:38 Now, let's work on these next ones because these are actually fun.

02:41 The square root of any irrational number is irrational.

02:45 The square root of any irrational number is irrational.

02:50 Okay, so we can say, write a negation.

02:53 So let's first write this as an if -then statement.

02:56 The square root of any irrational number is irrational.

03:00 So we'll say if, if x is not an element of the rational numbers, right? then the square root of x is not an element of the rational numbers.

03:19 Right? now we have something that we can negate.

03:22 Right? and whenever we negate an implication, right? when we negate an implication if we have, let me scroll up just a little bit here.

03:29 If we have an implication, p implies q, and we negate that.

03:34 And we can use the truth table to show this, but that's a waste of our time right now.

03:38 I think we have p and not q.

03:42 P and not q.

03:43 So that's how we're going to negate this.

03:45 And this is how we negate anything when we're talking proof by a contradiction.

03:49 We're going to say suppose not.

03:50 And that means when we say suppose not, suppose that our implication is false, which means we're going to rewrite our statement as let our premise be true.

03:59 And assume that our conclusion is not true.

04:02 That's what we're going to do.

04:04 So we're going to say, suppose not let x be irrational, which means it's not an element of the rational numbers, and then and let the square root of x be an element of the rational numbers.

04:25 Boom, and that's where we're going to start.

04:27 All right, so that's what we're going to say.

04:28 That's the negation.

04:30 Let x be not a rational number, and suppose that the square root of x is.

04:37 And that's where we're going to start.

04:38 So we'll say this is going to be our proof.

04:40 I always like to say suppose not.

04:43 Suppose not.

04:44 And then i would rewrite, let x this, right? so we would say, if the square root of x is rational, then we would start there.

04:56 So the square root of x equals sum a over b for for some a and b that are elements of the integer, elements of the integers, which are irrationals, where b doesn't equal 0, and b does not equal 0.

05:20 That's what it means by definition of irrational.

05:22 And of course, we can't divide by 0.

05:24 That's why we have to say that b can't equal 0.

05:27 And that's going to be really important when we have to multiply these together.

05:30 Okay, so if we have the square root of x equals a over b then x itself is going to be a squared over b squared right now we have a squared over b squared right so that means that it equals really which really we'll say uh instead of over b squared it equals a squared times one over b squared and that's important because the rationales and the integers are closed under multiplication and addition, which means that if i multiply any two rational numbers or any two integers, then i get another integer.

06:17 If i add any two integers, then i get another integer.

06:20 So right here, i'm multiplying an integer and an integer.

06:26 So, well, actually, that's not an integer.

06:28 They're rational.

06:29 The rationales are also closed.

06:31 Okay, so we have then, so we'll have a squared.

06:37 Times 1 over b squared is an element of the rationales because they're closed under addition and multiplication.

06:54 But that, okay, so then we can say, so x is an element of the rationales.

07:08 Then we can say, but x is not an element of the rationales, because that is what? but we started with, let x not be rational.

07:21 That's our contradiction.

07:23 So we would say the symbol varies from textbooks to textbooks, but this is the one that i learned for contradiction.

07:29 It's like two crossed hockey sticks, but x is not an element of the rationales, and that's our contradiction.

07:35 So therefore, what happens when we assume, we assumed, right, that the square root of x is not rational.

07:50 That means that the square root of x, or we assumed, i'm sorry, we assumed that the square root of x was rational, so that means that the square root of x has to be irrational, therefore.

08:03 The square root of x is not an element of the rational numbers, and that is the end of our proof.

08:11 This is the best part, the little box...

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