MATH 220 Homework 2 Solutions - Texas A&M University
MATH 220 Homework 2 Solutions
Mandatory Problems
Exercise 2.1.1
Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a | c, then a | (bm + cn). Proof. Let m and n be integers. Assume a | b and a | c. Then there exist integers j and k such that b = aj and c = ak. (We must show that there exists an integer l such that bm + cn = al.) Observe that
bm + cn = ajm + akn = a(jm + kn). Put l = jm + kn. Then l is an integer such that bm + cn = al, and therefore a | (bm + cn).
Exercise 2.1.4
For every integer n, n2 + n + 5 is odd. Proof. Let n be an integer. We consider two cases.
Case 1: n is even. If n is even, there is some integer k such that n = 2k. Then we have
n2 + n + 5 = 4k2 + 2k + 4 + 1 = 2(2k2 + k + 2) + 1. Put l = 2k2 + k + 1. Then l is an integer such that n2 + n + 5 = 2l + 1, and therefore n2 + n + 5 is odd.
Case 2: n is odd. If n is odd, then there is some integer k such that n = 2k + 1. Then we have
n2 + n + 5 = (4k2 + 4k + 1) + (2k + 1) + 5 = 2(2k2 + 3k + 3) + 1. Put l = 2k2 + 3k + 3. Then l is an integer such that n2 + n + 5 = 2l + 1, and therefore n2 + n + 5 is odd.
1
Exercise 2.1.11
Prove
that
for
all
real
numbers
x
and
y,
if
x
<
y,
then
x
<
x+y 2
<
y.
Proof.
Let
x
and
y
be
real
numbers
such that
x < y.
Then 2x = x + x < x + y, and thus
x<
x+y 2
.
Similarly,
x+y
<
y
+y
=
2y,
and
thus
x+y 2
<
y.
Combining
our
results,
we
obtain
x+y
x<
0. Observe that 0 (x - 1)2 = x2 - 2x + 1, so
2x x2 + 1 and thus
2 = 1 ? 2x 1 ? (x2 + 1) = x + 1
x
x
x
as desired.
Exercise 2.1.17
Prove
that
for
all
x, y
R+,
xy
x+y 2
,
with
equality
if
and
only
if
x
=
y.
Proof. Let x, y R+ be given. Observe that 0 (x - y)2 = x2 - 2xy + y2, and so 4xy
x2
+ 2xy + y2
= (x + y)2.
Thus
2xy
x + y,
and
therefore
xy
x+y 2
.
Assuming x = y, we obtain
xy
=
x2
=
x
=
2x
=
x
+
y .
2
2
Conversely,
assume
xy
=
x+y 2
.
Then
4xy
=
(x + y)2
=
x2
+ 2xy
+ y2,
and
rearranging
we
obtain
0 = x2 - 2xy + y2 = (x - y)2, so 0 = x - y and therefore x = y.
Exercise 2.1.21
Let a, b, c R with a = 0. Prove that the equation ax + b = c has a unique solution in R.
2
Proof.
Put
x0
=
c-b a
.
Then
x0
R
and
ax0 + b
= c,
so
the
equation
ax + b = c
has
a
solution.
If
now x1 is also a solution to the equation ax + b = c, then
0 = c - c = (ax0 + b) - (ax1 + b) = a(x0 - x1).
So x0 -x1 = 0, and thus x0 = x1. Therefore x0 is the unique solution to the equation ax+b = c.
Exercise 2.2.1
Let n be an integer. If n2 is even, then n is even.
Proof. Assume n is not even. Then n is odd, hence there is some k N such that n = 2k + 1. Thus n2 = (2k + 1)2 = 2(2k2 + 2k) + 1, and thus n2 is odd. Therefore, by contraposition, if n is even, then n2 is even.
Exercise 2.2.3
Prove that there are no integers m and n such that 8m + 26n = 1.
Proof. Assume, towards a contradiction, that there exist integers m and n such that 8m + 26n = 1. Then 2(4m + 13n) = 1, so 1 is even. But 1 is odd, a contradiction. Thus there are no integers m and n such that 8m + 26n = 1.
Exercise 2.2.10
Prove that for all integers m and n, if mn is even, then m is even or n is even. Proof. Assume that mn is even, and that m is not even. Then m is odd. Thus there exist integers j and k such that mn = 2j and m = 2k + 1. Then
2j = mn = 2kn + n, so n = 2(j - kn), where j - kn Z, and therefore n is even.
Exercise 2.2.13
Let x R. If for all > 0, |x| < , then x = 0.
Proof.
Assume, towards a contradiction, that x = 0.
Then
x 2
= 0, so
x 2
> 0.
Letting =
x 2
,
we obtain |x| < =
x 2
=
|x| 2
,
and
rearranging
yields
2
<
1.
This
is
a
contradiction,
and
therefore
x = 0.
3
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