Math 290 Lecture #7
§3.4-3.5: Introduction to Proofs, Part II
§3.4: Proof by Cases. Sometimes, we divide up a proof into cases, as determined by
the nature of the hypothesis.
This happens for instance when consider even and odd integers separately.
Or we may have two integers in the hypothesis and need to consider them having the
same parity, i.e., both are even or both are odd; or they could have the opposite parity,
i.e., one even and one odd.
Theorem 3.16. Let x, y ∈ Z. Then x and y are of the same parity if and only if x + y
is even.
Proof Strategy. With x and y being of the same parity we approach this with cases of x
and y are even, and x and y are odd.
Proof. Suppose that x and y are of the same parity.
Case 1: x and y are even. Then x = 2k and y = 2l for some k, l ∈ Z.
Now
x + y = 2k + 2l = 2(k + l).
Since k + l ∈ Z, we have that x + y is even.
Case 2. x and y are odd. Then x = 2k + 1 and y = 2l + 1 for some k, l ∈ Z.
Now
x + y = 2k + 1 + 2l + 1 = 2k + 2l + 2 = 2(k + l + 1).
Since k + l + 1 ∈ Z, we have that x + y is even.
This establishes the implication in the direction of left to right.: x and y of the same
parity implies x + y is even.
For the other implication, right to left, the strategy is to consider its contrapositive: x
and y of opposite parity implies x + y is odd.
Again we have two cases of x even and y odd, and of x odd and y even.
What you may notice here is both of these cases have one even integer and one odd
integer, and the proofs in each case will be almost exactly the same.
In this situation we can use the “Without Loss of Generality” (or WLOG for short).
WLOG, suppose that x is even and y is odd.
Then x = 2k and y = 2l + 1 for some k, l ∈ Z.
Now
x + y = 2k + 2l + 1 = 2(k + l) + 1.
Since k + l ∈ Z, we have that x + y is odd.
Do you see how that case of x odd and y even is almost exactly the same? This justifies
using WLOG because repeating essential the same argument twice is not economical.
Theorem 3.17. Let a, b ∈ Z. Then ab is even if and only if a is even or b is even.
Proof strategy. It seems easier to start with a is even or b is even and show that ab is
even. So we start with this.
Proof. Suppose that a is even or b is even.
As we will see, it does not matter which of a and b is even, so WLOG, we assume that
a is even.
Then a = 2k for some k ∈ Z.
Now
ab = (2k)b = 2(kb).
Since kb ∈ Z, we have that ab is even.
For the opposite implication, starting with ab is even does not tell us very much about a
or b.
So we will approach this by the contrapositive: if a is odd and b is odd, then ab is odd.
Suppose that a and b are odd.
Then a = 2k + 1 and b = 2l + 1 for some k, l ∈ Z.
Now
ab = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1.
Since 2kl + k + l ∈ Z, we have that ab is odd.
§3.5 Proof Evaluations. If we see a proof without a statement of the result, could
be recover the result?
Example. Here is a proof.
Suppose a is an even integer.
Then a = 2k for some k ∈ Z.
Now
5n + 1 = 5(2k) + 1 = 2(5k) + 1.
Since 5k is an integer, we have that 5n + 1 is odd.
What is this a proof of ?
Two correct answers are: If n is an even integer, then 5n + 1 is an odd integer, and if
5n + 1 is an even integer, then n is an odd integer.
We have already seen “proofs” that were not convincing; they really were not proofs
because they missed something or had some error in them.
Result. If x and y are integers of the same parity, then x − y is even.
Proof. Let x = 9 and y = 7. Then x − y = 9 − 7 = −2 which is even.
Is this a convincing proof? What is wrong with it?
We only showed that for the odd integers 9 and 7 we have 9 − 7 is even. What about
other pairs of odd integers?
We did not even consider the case of a pair of even integers.
So (I hope) we can see that this kind of argument is lacking; it fails to account for all
possible pairs of odd integers, and all pairs of even integers.
Here is a correct argument.
Proof. Suppose x and y are integers of the same parity.
Case 1. Assume x = 2k and y = 2l for k, l ∈ Z.
Then
x − y = 2k − 2l = 2(k − l).
Since k − l is an integer, then x − y is even.
Case 2. Assume x = 2k + 1 and y = 2l + 1 for some k, l ∈ Z.
Then
x − y = 2k + 1 − (2l + 1) = 2k + 1 − 2l − 1 = 2k − 2l = 2(k − l).
Since k − l ∈ Z, then x − y is even.
Result. If m is an even integer and n is an odd integer, then 3m + 5n is an odd integer.
Proof. Suppose m is an even integer and n is an odd integer.
Then m = 2k and n = 2k + 1 for some k ∈ Z.
Now
3m + 5n = 3(2k) + 5(2k + 1) = 6k + 10k + 5 = 16k + 4 + 1 = 2(8k + 2) + 1.
Since 8k + 2 ∈ Z, then 3m + 5n is even.
Is this a convincing proof? What is wrong with it?
We related m and n through the integer k, but no such relationship was assumed in the
hypothesis.
We did not give a proof in the most general case of m an even integer and n an odd
integer.
Here is a correct argument where we do not mistakenly relate m and n.
Proof. Suppose m is even and n is odd: we have m = 2k and n = 2l + 1 for some k, l ∈ Z.
Then
3m + 5n = 3(2k) + 5(2l + 1) = 6k + 10l + 5 = 6k + 10l + 4 + 1 = 2(3k + 5l + 2) + 1.
Since 3k + 5l + 2 ∈ Z, then 3m + 5n is odd.