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### Contrapositive, Converse, Inverse

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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.
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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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Associated Topics:
High School Discrete Mathematics
High School Logic

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