HomeWork Answers
Section 1.6
1. Prove: The sum of two odd integers is even.
Proof:
Let:
a = 2k + 1
b = 2l + 1
Then:
a + b = 2k + 1 + 2
l + 1 = 2(k + l +1)  EVEN
2. Prove: The sum of two even integers is even.
Proof:
Let:
a = 2k
b = 2l
Then:
a + b = 2k + 2
l = 2(k + l)  EVEN
3. Prove: The square of an even integer is even.
Proof:
Let:
a = 2k
Then:
a2 = (2k)2 = 4k2 = 2(2k2)
 EVEN
6. Prove: The product of two odd integers is odd.
Proof:
Let:
a = 2k + 1
b = 2l + 1
Then:
ab = (2k + 1)*(2
l + 1) = 4kl + 2k + 2l +1 =
2(2kl + k + l) + 1
 ODD
7. Prove: Every odd integer is the difference of two squares.
Discussion:
1
Often it is good to gain an intuitive grasp by examining examples. This can
serve to give us confidence in what we seek to prove, but also may lead us to find
patterns that can be useful in the proof. Let’s see look at examples of odd integers
and their respective differences of squares.
x
1
3
5
7
9
11
-1
-3
-5
-7
-9
-11
m2
1
4
9
16
25
36
0
1
4
9
16
25
n2
0
1
4
9
16
25
1
4
9
16
25
36
m
1
2
3
4
5
6
0
1
2
3
4
5
n
0
1
2
3
4
5
1
2
3
4
5
6
Two important observations come out of this exploration. 1st, that for positive
odd integers, m = n + 1, and for negative odd integers, m = n – 1. 2nd, that x = 2n
+ 1, for positive integers and x = -2n + 1, for negative ones. We add to that some
observations from high school algebra. Since
x = m2 – n2,
by factorization
x = (m + n)*(m – n) =
(n + 1 + n)*(n + 1 – n) =
(2n + 1)*(1) =
2n + 1
In other words, if
x = 2n + 1,
then
x = (2n + 1)*1 = (n + n + 1)*(n + 1 – n) = (n + 1) 2 - n2.
What is x is a negative odd integer? We see that case is covered if we let m = n – 1.
Then
x = (n – 1 + n)*(n – 1 – n) = (2n – 1)*(–1) = – (2n – 1) = –2n + 1.
The proof, then, is just a summary of the above.
Proof:
There are two cases – x is a positive odd integer or x is a negative odd
2
integer.
Case 1 – let:
x = 2k + 1
Then, by the above, x =
(2k + 1)*(1) =
k2 + 2k + 1 – k2 =
(k + 1)2 – k2 =
m2 – n2
Case 2 – let:
x = – (2k + 1)
Then, by the above, x =
–2k – 1 =
k2 – 2k – 1 – k2 =
(k – 1)2 – k2 =
m2 – n2
8. Prove: If n is a perfect square, then n+2 is not.
Proof: By contradiction
WOLG, assume n and n+2 are both nonnegative.
Suppose:
n and n+2 are both perfect squares.
Let:
n = k2, k  0
n+2 = l2, l  0
Then:
l2 – k2 = 2
and, since l2 > k2 ,
Let:
l = k + a, a > 0.
Suppose:
a=1
Then:
l>k
l2 – k2 = (k + 1)2 – k2 = k2 + 2k + 1 – k2 = 2k + 1  ODD  2 
So:
a > 1, i.e., a  2, and a2 > 2
Then:
l2 – k2 = (k + a)2 – k2 = k2+ 2ka + a2 – k2 = 2ka + a2 > 2 
Example: if a = 2,
l2 – k2 = (k + 2)2 – k2 = k2 + 4k + 4 – k2 = 4k + 4 > 2 
Conclusion:
Our hypothesis that n and n+2 are both perfect squares fails. So the
3
theorem is proven.
11. Disprove: The product of two irrational numbers is always irrational.
Disproof: By counterexample:
We know 2 is irrational.
And 2 * 2 = 2, which is rational.
16. Prove: If mn is even, where m and n are integers, then either m is even or n is.
Proof: By contradiction:
Suppose:
Both m an n are odd integers.
Let:
m = 2k + 1
n = 2l + 1
Then:
mn = (2k + 1)*(2 l + 1) =
4kl + 2k + 2l + 1 =
2(2kl + k + l) + 1 =  ODD 
Conclusion:
Our hypothesis that both m and n are odd fails. So the theorem is
proven.
38. Find counterexample: Every positive integer can be written as the sum of
squares of three integers.
Discussion:
Try some examples and look for a pattern.
x
1
2
3
4
5
6
7
sum
0 +
0 +
1 +
0 +
0 +
1 +
? +
of 3 squares
0 + 1
1 + 1
1 + 1
0 + 4
1 + 4
1 + 4
? + 4
Conclusion:
The only squares < 7 are 0, 1, and 4. 0 and 1 by themselves cannot add
up to 7, if we sum 3 of them. So, at least one of the addends must be 4. On
the other hand, only one addend can be 4, since 4+4 > 7. So we need the
other two addends to sum up to 3. But 0 and 1 by themselves cannot.
4