Analysis Homework #1
Solutions
1. Determine max A when A = {y ∈ R : y = −3x2 + 2x − 1 for some x ∈ R}.
• Completing the square, we find that every element of A has the form
(
)
(
)2
2x 1 1 1
1
2
2
2
−3x + 2x − 1 = −3 x −
+ − +
= −3 x −
− .
3
9 9 3
3
3
Using this fact, it is now easy to see that −2/3 is the largest element of A.
n−1
2. Show that the set B = { 2n+1
: n ∈ N} has no maximum.
• To see that B has no maximum, suppose x =
y=
n−1
2n+1
is the maximum and let
(n + 1) − 1
n
=
.
2(n + 1) + 1
2n + 3
Then y is an element of B, and it is actually larger than x because
n−1
n
<
2n + 1
2n + 3
⇐⇒
2n2 + n − 3 < 2n2 + n
⇐⇒
−3 < 0.
As no element of B can be larger than the maximum of B, this is a contradiction.
n−1
: n ∈ N} is such that sup B = 1/2.
3. Show that the set B = { 2n+1
• To show that 1/2 is an upper bound of the given set, we note that
n−1
1
≤
2n + 1
2
⇐⇒
2n − 2 ≤ 2n + 1
⇐⇒
−2 ≤ 1.
Since the rightmost inequality is true, the leftmost one must also be true.
• To show that 1/2 is the least upper bound, we show that no number x < 1/2 is an
upper bound. Let us then fix some x < 1/2 and try to find an element of B which is
n−1
bigger than x. Since every element of B has the form 2n+1
, we need to ensure that
n−1
>x
2n + 1
⇐⇒
n − 1 > 2nx + x
⇐⇒
(1 − 2x)n > x + 1.
Noting that 1 − 2x is positive whenever x < 1/2, we end up with the inequality
n>
x+1
.
1 − 2x
According to one of our theorems, we can always find an integer n that satisfies this
n−1
inequality. Then, our computation above shows that 2n+1
is an element of B which is
bigger than x. This also means that x is not an upper bound of B, as needed.
4. Show that the set C = {x2 : x ∈ R} has no upper bound.
• If y ∈ R is an upper bound of C, then y ≥ y 2 and also y ≥ 22 > 1. Multiplying the
last inequality by the positive number y, we now get y 2 > y, a contradiction.