Contrapositive, Converse, Inverse
Date: 06/10/99 at 19:48:55 From: Hollye Goode Subject: Mathematical reasoning (logic) Let m and n be whole numbers, and consider the statement p implies q given by "if m + n is even, then m and n are even." A) Express the contrapositive, the converse and the inverse of the given conditional. B) For the statements that are true, give a proof. C) For the statements that are false, give a counterexample. I have part A (I think) but I'm having trouble deciding which statements are true and which are false, and I'm completely lost on the proofs.
Date: 06/11/99 at 18:31:46 From: Doctor Kate Subject: Re: Mathematical reasoning (logic) Hollye, I'll give you what I got for the first part, to see if it's the same as what you got. First, though, here's what my "p" and "q" are: p is "m + n is even" q is "m and n are even" ~p is "m + n is odd" ("~p" means "NOT p") ~q is "either m or n is odd" A. Contrapositive (if ~q then ~p): "If either m or n is odd, then m + n is odd." B. Converse (if q then p): "If m and n are even, then m + n is even." C. Inverse (if ~p then ~q): "If m + n is odd, then either m or n is odd." To check out which of these are true, it's best to experiment a little. Try some numbers. Let's look at and pick some numbers where m or n is odd: 2 and 3 3 and 7 1 and 8 Notice that I tried to pick a variety of numbers - sometimes both odd, sometimes only one. That is because the opposite of "m and n are even" is "at least one of m or n is odd, and maybe both are." You can figure that out by imagining all sorts of things that don't satisfy "m and n are even." It could be really false (both m and n are not even) or just a bit false (only n is not even or only m is not even). Anyway, let's take a look at these numbers. Is 2 + 3 odd? Yes. Is 3 + 7 odd? That's 10... no, it's not. Wait, statement A says 3 + 7 WOULD be odd. This is a counterexample. Remember that a statement like "<BLAH> is always true" can be proven false by just one example of when <BLAH> could be false. If I claim all dogs are black, all you have to do is bring me a Dalmation, and I am wrong, even if a lot of dogs are black. Statement A is claiming that ALL the time, if one or both of n or m is odd, n+m is ALWAYS odd. But look, we found an example where it isn't. So statement A is false. Now you try a few numbers for statement B. After a while, you'll get tired of checking them out because you won't find a counterexample (but you should still look, before you try to prove something true). So let's try to prove this one. Here it is again: "If m and n are even, m + n is even." What does it mean for a number to be even? It's divisible by two. So here's a useful way to think of that: If m is even, then m = 2*r for some number r. Think about that for a minute and you'll agree it's a pretty good definition of evenness. It's useful too. Let's translate what we're given and what we want to prove using this definition: We are given: m = 2*r for some number r n = 2*s for some number s We want to prove: m + n = 2*t for some number t Check this out for yourself - play around a little; try to work it out. Setting up the question properly is half of doing a proof. This is just re-stating a question in a way that might seem easier to deal with. Notice that I used three different numbers r, s and t. There's no reason for them to be the same is there? It probably ISN'T true that r = s. So let's try to prove the statement. Let's "investigate" m + n by using the given information: m + n = 2*r + 2*s for some numbers r and s, by the "given" information = 2*(r+ s) by "factoring out" the 2. Now let t = r + s. And we have proven that m + n = 2*t for some number t (in fact, the number t turns out to be r + s). That's it - it's not a long proof at all - all the work is in setting it up properly. Now why don't you give the others a try. I'll give you a sketch of what I've found to compare to. Hint: for "odd" use the following definition: If x is odd, then x = 2*r + 1 for some number r. Here's what I got. Statement C is true. If m + n is odd, m + n = 2*r + 1 for some number r. Suppose m is even. Then m = 2*s for some number s. Then 2*s + n = 2*r + 1 (by substituting in the equation above) Then n = 2*r - 2*s + 1 (by subtracting 2*s from both sides) Then n = 2*(r - s) + 1 (by factoring out the 2) So n is equal to 2*t + 1 for some number t. So n is odd. Similarly, if I suppose n is even, then I can show m is odd. (Do it!) So at least one of n or m is odd. If you didn't do it this way, don't worry. There are lots of ways to do it. If you did it by contradiction (suppose both n and m are even) you could have used statement B to prove it! In fact, statements B and C are equivalent. They "mean" exactly the same thing. Think about that for a while. Finally, the original statement is false. I'll let you look for a counterexample. Remember, try lots of different sorts of examples. Hint: What's the simplest example you could try? - Doctor Kate, The Math Forum http://mathforum.org/dr.math/
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