The Basics of Proofs Spring 2010 BYU Math Department Today's - PowerPoint PPT Presentation

The Basics of Proofs Spring 2010 BYU Math Department

Today's Topics 1)Using definitions, theorems and laws 2)Using a counter-example 3)Proof by Contradiction 4)Proof by Induction

General Rule of Thumb When you see a problem, always ask the following questions: 1)What do they want? 2)What do I know? 3)How can I use what I know to get what they want?

Using Definitions, Theorems, and Laws Proofs are very different from the traditional math problem. Often the problem won't tell you all the information you need. This is why it's important to know the definitions, theorems, and laws listed throughout the section, chapter, and book.

Let's try an example First, you need some definitions. These may seem obvious, but pretend that you've never been taught this stuff before. Prove: 1 + 1 = 2 1 + 1 + 3 = 5 2 + 3 = 5 Let's start by asking the basic questions 1)What do they want? 2)What do I know? 3)How can I use what I know to get what they want?

Example 1, cont. What do they want? How do I use what I know to get what they 1 + 1 + 3 = 5 want? What do I know? 1 + 1 = 2 Since 1 + 1 = 2 2 + 3 = 5 Then 1 + 1 + 3 = 2 + 3 And we know that 2 + 3 = 5 Therefore 1 + 1 + 3 = 5. => QED

Quick Gear Change! Quick Gear Change! You shouldn't (and your professors won't!) use standard English to write out the steps of your proofs. We will use notation instead. => “implies that” Ǝ “there exists” Ʉ “for all” ϶ or s.t “such that” . . . “therefore” QED quod erat demonstrandum (end of proof)

So let's 'fix' our last proof Since 1 + 1 = 2 1 + 1 = 2 Then => 1 + 1 + 3 = 2 + 3 1 + 1 + 3 = 2 + 3 becomes 2 + 3 = 5 And we know that =>1 + 1 + 3 = 5. 2 + 3 = 5 QED Therefore 1 + 1 + 3 = 5. QED

Let's try something more realistic Suppose the following: Jerry needs help if and only if Jerry is a rudy pants. Jerry is loony if Jerry needs help. Jerry is a rudy pants. Prove: Jerry is loony.

Example 2, cont. What do they want? How do I use what I know to get what they Prove that Jerry is loony want? What do I know? Jerry is a rudy pants Jerry needs help if and only => Jerry needs help if Jerry is a rudy pants. => Jerry is loony QED Jerry is loony if Jerry needs help. Jerry is a rudy pants.

Let's try something more math-like Prove: If f(x) is any kind of even function, then f(x) is not a one-to-one function

Example 3, cont. How is this problem different? All of the information you need has not been given to you in the stating of the problem. They assume that you know your definitions of even/odd/one-to-one functions. Quick Review! If f(x) is an even function, then f(-x) = f(x) If f(x) is a one-to-one function, then every value f(x) can be reached by one and only one value x.

Example 3, cont. What do they want? How do I use what I know to get what they Prove that f(x) is not one-to-one want? What do I know? f(x) is even f(x) is even => f(-x) = f(x) => Ǝa,b ϶ f(a) = f(b) If f(x) is even, then =>There is more than one f(-x) = f(x) corresponding x for every f(x) If a function is one-to-one, then ... f(x) is not one-to-one. for every f(x) there is one QED and only one corresponding x.

Let's try one more example Prove: If y = x, and y = x 2 then y = 0 or 1 Any Volunteers?

Example 4, cont. What do they want? How do I use what I know to get what they want? Prove that y = 0 or 1 y = x What do I know? y = x 2 => x 2 = x y = x => x 2 - x = 0 y = x 2 => x(x - 1) = 0 => x = 0, 1 If a = b, and b = c y = x then a = c ... y = 0, 1 QED

Tips and Tricks for Future Problems If you just can't seem to solve the problem... 1)Make sure you found all possible definitions, theorems, and laws that are associated with the problem. 2)Start writing something! – Most of the time you won't see how to prove it just by thinking about it. 3)Try to understand the proof conceptually. – Sometimes understanding why the proof is true from another angle opens your mind to possible solutions.

Using a Counter-example Sometimes you'll be given a problem where you need to disprove something. The method to solving these problems is much simpler. All you need is a counter-example.

The Indian Example I have a close friend who visited a nearby Indian reserve and insisted that all Native Americans walk in single file. I asked him how he knew that. He said that the one that he saw was walking in single file.

Learning from the Indian Example Why was my friend's conclusion absurd? He used only one case to say something very general about Native Americans. What if he had told me that he knew that not all Native Americans walked in single file because he saw two walking side by side? His statement would be accurate, right?

Let's try another example Disprove: If a and b are any odd numbers, then a + b is always an odd number.

Example 5, cont. What do they want? How do I use what I know to get what they want? Disprove that a + b is always odd. a and b are odd numbers What do I know? Let a = 1, b = 3 => a + b = 4 a and b are any odd numbers => 4 is even => a + b is even ... a + b is not always odd QED

Don't forget! You can only DISPROVE something with an example. It is NOT enough to just use an example to PROVE something generally.

Proof by Contradiction Let's suppose that you could prove that something was never false. That would mean that it must always be true, right? That is the premise of proofs by contradiction. We will assume that the conclusion is false and show that it leads to a contradiction.

Let's try another example Prove: 1/2 is an irrational number 2 2 is even => x is even Hint: if x

Example 6, cont. How do I use what I know What do they want? to get what they want? 1/2 is irrational Show 2 1/2 is rational Assume 2 => Ǝ p/q ϶ p/q = 2 1/2 What do I know? 2 /q 2 = 2 => p 2 = 2q 2 => p Ʉk, 2k is an even number An irrational number cannot 2 is an even number => 2q be represented by an 2 is an even number => p irreducible fraction such => p is an even number as p/q. => Ǝm, ϶ p = 2m 2 = 4m2 => p 2 is even => x is even x 2 = 2q 2 => 4m 2 = q 2 => 2m => q 2 is even => q is even => p/q is irreducible ... Our assumption is always false 1/2 is irrational => 2 QED

Let's try one more example Prove: There do not exist integers m and n such that 4m + 6n = 9.

Example 7, cont. What do they want? How do I use what I know to get what they want? 4m + 6n ≠ 9 for any m and n Assume that Ǝ m, n s.t. 4m + 6n = 9 What do I know? => 2(2m + 3n) = 9 => 2(2m + 3n) is even => an even number = 9 Not much... => which is false ... Our assumption is always false => 4m + 6n ≠ 9 for any m and n

Proofs by Induction Let's suppose you wanted to prove that a pattern works for any possible case up to infinity. For example, how could you prove that 1 + 2 + … + n = n(n+1)/2 regardless of what n was? What if I could prove that it worked for n = 1? Is that enough?

Proofs by Induction, cont. 1 + 2 + … + n = n(n+1)/2 What if we could prove it was true for 1 and 2? Is that enough? What if we could prove it was true for any random value as well as the next one? Is that enough? We need to prove both.

The three basic steps 1. Prove that it works for a base case 2. Assume that it works for a random value k 3. Prove that it works for the value k + 1 You will always use these three steps when working with induction.

Let's solve our problem Prove: 1 + 2 + … + n = n(n+1)/2 for all n FYI: If a proof asks you to prove it for all values of n or similar, then induction will be the method to use 99% of the time.

Example 8, cont. What do they want? How do I use what I know to get what they want? 1 + 2 + … + n = n(n+1)/2 Show that it works for n = 1 => 1 = 1(1+1)/2 What do I know? => 1 = 1(2)/2 = 1(1) = 1 TRUE Sounds like an induction Assume that the pattern works for problem. n = k => 1 + 2 + … + k = k(k+1)/2 Prove that it works for n = k+1 => 1 + 2 + … + k + k+1 = (k+1)(k+2)/2 1 + 2 + … k = k(k+1)/2 => k(k+1)/2 + k + 1 = (k+1)(k+2)/2 => (k+1)(k/2 + 1) = (k+1)(k+2)/2 => (k+1)(k+2)/2 = (k+1)(k+2)/2 TRUE ... 1 + 2 + … + n = n(n+1)/2 QED

Let's do one more hard problem Prove: 2 * 6 * 10 * … * (4n – 2) = (2n)!/n! Any volunteers?!