CSCE 235
Handout
11: Examples of Proofs
February 4, 2004
1. Prove that the product of two odd integers is an odd integer.
First, let the first odd integer be a = 2k + 1 such that k is an integer, and the second odd integer be b = 2m + 1 such that m is an integer. Now, the product of such two odd integers is then ab = (2k + 1)( 2m + 1) = (2k)(2m) + 2k + 2m + 1 = 4km + 2k + 2m + 1 = 2(2km + k + m) + 1. Since 2(2km + k + m) is always even, we know that 2(2km + k + m) + 1 is always odd! Thus, we have proved that ab is an odd integer. Therefore, we have proved that the product of two odd integers is an odd integer.
2. Prove that 
is irrational. (Hint: Very similar to the example that I
gave in class that proves 
is irrational.)
Our approach here
is to use proof by contradiction. That
is, we first assume that 
is rational. Thus, assume that there is a number x
= 
such that x = 
such that both m and
n are integers, and 
. And since 
is a rational number,
we know that 
is a ratio such that m
is not divisible by n; i.e., m and n do not have
common factors. Now, we have:
So, we can
substitute 
into x = 
so that we have 
. However, we can
reduce the above equation to 
. But by assuming
that x = 
is rational, 
cannot be further
reduced (as stated earlier in our proof).
But we have just shown that we have reduced 
to 
. Thus, we have
encountered a contradiction. Therefore,
is irrational.
3. Prove that the sum of three consecutive integers is divisible by 3.
First, let the first integer be k, the second integer be k + 1, and the third integer be k + 2. Thus, we have three consecutive integers. The sum of these three integers is k + (k + 1) + (k + 2) = 3k + 3 = 3(k + 1). And since 3(k + 1) is always divisible by 3, we have proved that the sum of three consecutive integers is divisible by 3.
4. Prove
that 
is divisible by 3 for
all 
N.
To show that 
is divisible by 3, we consider three cases: n =
3k, n = 3k+1, and n = 3k+2, where 
N. The three cases cover all possible numbers in N. So, we will show the proof by cases.
Case (i):
when n = 3k: 
= 
. Since the number is a multiple of 3, it is divisible by 3. Thus, we have shown that for n = 3k,
is divisible by 3.
Case (ii):
when n = 3k+1: 
=
.
Since the number is a multiple of 3, it is divisible
by 3. Thus, we have shown that for n
= 3k+1, 
is divisible by 3.
Case (iii):
when n = 3k+2: 
=
.
Since the number is a multiple of 3, it is divisible
by 3. Thus, we have shown that for n
= 3k+2, 
is divisible by 3.
Therefore, we have shown that for all cases, 
is divisible by 3.