PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM
General Comments
: These are practice problems. You should regard each indi
vidual problem as being a plausible exam problem. Certainly there are too many
problems here to make one hourlong exam (roughly twice as many).
The exam will be closedbook and calculators will not be permitted. I would
urge you not to use calculators or software when doing the practice problems – that
would not be such good practice. When in doubt, you should justify your answers
by correct logical reasoning (brief explanations are often suﬃcient).
1) Let
a,b
∈
R
.
a) Suppose
a
and
b
are both positive. Show that
a < b
⇐⇒
a
2
< b
2
.
b) Show that
a < b
⇐⇒
a
3
< b
3
.
c) Suggest a generalization of the ﬁrst two parts to
n
th powers, for an arbitrary
positive integer
n
. You need not prove it.
First Solution:
a) Suppose
a,b >
0. We want to show that
a
−
b <
0
⇐⇒
a
2
−
b
2
<
0. In general,
given nonzero numbers
A
and
B
, to show that
A
and
B
have the same sign it is
enough to show that
A
B
is positive. As
a
2
−
b
2
= (
a
−
b
)(
a
+
b
), we have
a
2
−
b
2
a
−
b
=
a
+
b,
which is indeed positive since
a
and
b
are both positive.
b) Now suppose
a
and
b
are arbitrary but both zero (if
a
=
b
= 0, neither
a
−
b
nor
a
3
−
b
3
is negative). As above, we have
a
3
−
b
3
a
−
b
=
(
a
−
b
)(
a
2
+
ab
+
b
2
)
a
−
b
=
a
2
+
ab
+
b
2
,
so it’s enough to show that
a
2
+
ab
+
b
2
>
0. This is clear if
a
and
b
have the same
sign: all the terms are nonnegative and at least one is positive. If
a
and
b
have
opposite sign, then
a
2
+
ab
+
b
2
=

a

2
− 
ab

+

b

2
≥ 
a

2
−
2

ab

+

b

2
= (

a
 − 
b

)
2
≥
0
,
and is strictly positive unless
b
=
−
a
, in which case
a
2
+
ab
+
b
2
=
a
2
>
0. c)
The desired generalization is that if
f
(
x
) =
x
n
and
n
is even, then for all
a,b >
0,
a < b
⇐⇒
a
n
< b
n
, whereas if
n
is odd, then for all
a,b
∈
R
,
a < b
⇐⇒
a
n
< b
n
.
Remark: To prove this by the current factorization method involves the identity
a
n
−
b
n
= (
a
−
b
)(
a
n
−
1
+
a
n
−
2
b
+
...
+
ab
n
−
2
+
b
n
−
1
.
When
a
and
b
are both positive, it is clear that
a
n
−
1
+
a
n
−
2
b
+
...
+
ab
n
−
2
+
b
n
−
1
>
0
,
c
⃝
Pete L. Clark, 2011.
1