We will show that for any integer n, the number 3n^2 +5n is even. To do this,
consider two cases :
Case 1. n is even.
Then n = 2k for some k ∈ Z. Therefore 3n^2+5n =
3(2k)^2+5(2k) = 12k^2+10k =
2(6k^2 + 5k). Since 6k^2 + 5k ∈ Z, the number 3n^2 + 5n
Case 2. n is odd.
Then n = 2k + 1 for some k ∈ Z, and 3n^2 + 5n = 3(2k +
1)^2 + 5(2k + 1) =
12k^2+12k+3+10k+5 = 12k^2+22k+8 = 2(6k^2+11k+4). Since 6k^2+11k+4
the number 3n^2 + 5niseven.
Since 3n^2 + 5n is never odd, the implication follows. (This is a vacuous
(b) If n is even, then 3n^2-2n -5 is odd.
Suppose n is even. Then n = 2k for some k ∈Z.
Since 6k^2-2k-3 ∈ Z, the number 3n^2-2n-5 is odd.
(This is a direct proof.)
(c) If 7n^2 + 4 is even, then n is even.
We will prove this statement by contrapositive. Suppose n is odd. Then n = 2k
for some k ∈ Z. Therefore 7n^2 + 4 = 7(2k + 1)^2 + 4 =
7(4k^2 + 4k + 1) + 4 =
28k^2 +28k +11 = 2(14k^2 +14k +5)+1. Since 14k^2 +14k +5 ∈
Z, 7n^2 +4 is odd.
(d) If n-5m is odd, then n and m are of opposite parity.
We will prove the statement by contrapositive. Thus we will prove that if n
are of the same parity, then n-5m is even.
Case 1. n and m are both even.
Then n = 2k and m = 2l for some k,l ∈ Z. Therefore
n-5m = 2k-5(2l) =
2k-10l = 2(k-5l). Since k-5l ∈ Z, the number n-5m is
Case 1. n and m are both odd.
Then n = 2k + 1 and m = 2l + 1 for some k,l ∈ Z. Then
n -5m = 2k + 1 -
5(2l +1) = 2k +1-10l-5 = 2k-10l-4 = 2(k-5l-2). Since k-5l-2 ∈
the number n-5m is even.
(e) If 5 | (n-1), then 5 | (n^3 + n-2).
Suppose 5 | (n-1). Then (mod 5). Therefore
This implies that 5 | (n^3 + n-2). (This is a direct proof.)
Another proof: Suppose 5 | (n-1). Then n-1 = 5k for some k
∈ Z. Therefore
n = 5k+1, and we have n^3 +n-2 = (5k+1)^3 +(5k+1)-2 = 125k^3 +75k2 +15k+
1+5k+1-2 = 125k^3+75k^2+20k = 5(25k^3+15k^2+4k). Since 25k^3+15k^2+4k
we have 5 | n^3 + n-2. (This is also a direct proof.)
(f) 3 | mn if and only if 3 | m or 3 | n
We have two implications to prove.
-> Suppose 3 | mn. Show that 3 | m or 3 | n.
We will prove this statement by contrapositive. Suppose
m = 3k + c and n = 3l + d for some k, l, c, d ∈ Z where c
and d are equal to
either 1 or 2. We have
Then cd = 4, and mn = 3(3kl+2k +2l)+4 = 3(3kl+2k +2l+1)+1. Thus
<-Suppose 3 | m or 3 | n. Show that 3 | mn.
We will prove this statement directly. Without loss of generality, suppose 3
Then m = 3k for some k ∈ Z. Therefore mn = 3kn is
divisible by 3.
3. The number log32 is irrational.
Suppose log32 is rational. Then log32 = m/n for some
m,n ∈ Z, n > 0. Therefore 3m/n = 2,
so 3m = 2n. Since n > 0, 3m = 2n > 1. We know that
3m is odd since it is a
of odd integers (or you can prove a little lemma here that says, If a is odd and
m is a positive integer , then am is odd.). But we also know that 2n is even since it is
of even integers (or you can prove another similar lemma if you are not
is a contradiction, since an odd number cannot be equal to an even number.
log32 is irrational. (This is a proof by contradiction.)
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