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
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PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM
and this is enough to show the result when
n
is even. However, when
n
is odd we
need to establish the above inequality for all
a,b
which are not both zero, and oﬀ
the top of my head this does not seem so straightforward. It is worth pointing out
however that the result is actually equivalent to this inequality, so the alternate
solution below gives an indirect proof of it.
Second Solution:
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 Spring '08
 Staff
 Math, Limit of a sequence

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