Download MATH 8 MIDTERM SOLUTIONS MAY 9, 2005 (1

MATH 8 MIDTERM SOLUTIONS
MAY 9, 2005
(1) Suppose that x, y ∈ R. Either prove or give a counterexample to each of the following
statements:
(a) If x is rational and y is irrational, then x + y is irrational.
Solution: Suppose that x + y were rational, with x + y = a/b. Let x = c/d. Then
a c
ad − bc
a
y = −x= − =
,
b
b d
bd
and so y is rational also. This contradicts our initial assumption that y is irrational. Hence
it follows that x + y is irrational.
(b) If x is rational and y is rational, then x + y is rational.
Solution: Suppose that x = a/b and y = m/n. Then
an + bm
a m
x+y = +
=
,
b
n
bn
and so x + y is rational.
(c) If x is irrational and y is irrational, then x + y is irrational.
√
√
Solution: Suppose that x = 2 and y = − 2. Then x and y are irrational. However
√
√
x + y = 2 − 2 = 0,
and this is a rational number.
(2) Let P (n) denote the statement
1
1 + 2 + 3 + · · · + n = n(n + 1).
2
Prove by mathematical induction that P (n) is true for all integers n ≥ 1.
Solution: We first observe that 1 = 21 · 1 · 2, and so P (1) is true.
Let k be a positive integer for which P (k) is true. Then
1
1 + 2 + · · · + k + (k + 1) = k(k + 1) + (k + 1)
(by our inductive hypothesis)
2
1
= (k + 1)(k + 2).
2
Hence, if P (k) is true, then so is P (k + 1). It therefore follows by induction that P (n) is
true for all integes n ≥ 1.
(3) Prove by mathematical induction that, for all integers n ≥ 0, 72n − 5n is divisible by
11.
Solution: Let P (n) denote the statement ‘72n − 5n is divisible by 11’.
1
2
MATH 8 MIDTERM SOLUTIONS MAY 9, 2005
Observe that 72·0 − 50 = 1 − 1 = 0. Since 0 is divisible by 11, it follows that P (0) is true.
Now let k ≥ 0 be an integer such that P (k) is true. Then
72(k+1) − 5k+1 = 72k · 72 − 5k · 5
= 72k (44 + 5) − 5k · 5
= 44 · 72k + 5(72k − 5k ).
Since 44 · 72k is divisible by 11, as is (by our inductive hypothesis) 5(72k − 5k ), it follows that
72(k+1) − 5k+1 is also divisible by 11.
Hence, if P (k) is true, then so is P (k + 1), and so it follows by induction that P (n) is true
for all integers n ≥ 0.
(4) Let x be an arbitrary integer.
(i) Prove that x(x + 1) is always even.
Solution: If x is even, then x = 2k for some integer k. This implies that
x(x + 1) = 2k(2k + 1).
Since this is a multiple of 2, it is even.
If, on the other hand x is odd, then x = 2k + 1 for some integer k. This implies that
x(x + 1) = (2k + 1)(2k + 2) = 2(2k + 1)(k + 1),
which is again a multiple of 2, and so is even.
(ii) Prove that, if x is not even, then x2 − 1 is divisible by 8.
Solution: If x is not even, then x = 2k + 1 for some integer k. Then
x2 − 1 = (2k + 1)2 − 1 = (4k 2 + 4k + 1) − 1 = 4k 2 + 4k = 4k(k + 1).
From (i), we know that k(k + 1) is even, and so k(k + 1) = 2m, say. Then
x2 − 1 = 4k(k + 1) = 4 · 2m = 8m,
and so x2 − 1 is divisible by 8.