MAS221 Analysis (Semester 1)– Solutions to Problems 1 to 7.
1. (a) c = 12 (a + b) is rational, and satisfies a < c < b, for c − a = b − c =
1
(b − a) > 0.
2
(b) One approach (it is not the only one) is to obtain the numbers
c1 , c2 , . . . , cn−1 by cj = a + nj (b − a) for 1 ≤ j ≤ n − 1. Clearly
(b − a) > 0. Note that in this
cj − a ≥ 0, while b − cj = n−j
n
construction, the numbers c1 , c2 , . . . , cn−1 are equally spaced with
cj − cj−1 = (b − a)/n for 1 ≤ j ≤ n − 1.
(c) Suppose there are only a finite number N (say) of distinct rational
numbers lying between a and b. Then we obtain a contradiction
with (b), by taking n = N + 2.
√
√
2. If x is rational, we may write x = ab where a ∈ Z+ and b ∈ N. But
2
then x = ab2 is also rational, and that gives the desired contradiction.
√
√
√
√
( p + q)2 = p + q + 2 pq. Now pq is not a perfect square, so pq
√
√ 2
is irrational by Theorem 1.2.3. It follows that ( p + q) is irrational,
and hence so is its square root, by the first part of this problem.
√ √2
3. The answer is “yes”, and here is a “non-constructive” proof.
√ 2 is
either rational or irrational. If√ it is rational choose a = b = 2, and if
√
√ 2
and
b
=
it is irrational
choose
a
=
2
2. [In fact it can be shown
√ √2
that 2 is irrational, but this needs some very advanced techniques.]
4. The two cases (i) and (ii) are both easy, as the inequality is an equality.
For case (iii), we have
|a + b| = a − |b| ≤ |a| + |b|.
For case (iv), a ≥ 0, b < 0 and |a| < |b|, we have similarly
|a + b| = |b| − a ≤ |a| + |b|.
The other two cases, are as in (iii) and (iv), but with the roles of a and
b interchanged.
5. It works when n = 1 as both sides are equal. Now suppose the inequality is true for some n ∈ N. Then
(1 + x)n+1 =
≥
=
≥
1
(1 + x)n (1 + x)
(1 + nx)(1 + x)
1 + (n + 1)x + nx2
1 + (n + 1)x,
and the result is proved by induction. When x > 0, the binomial
theorem yields
1
1
(1 + x)n = 1 + nx + n(n − 1)x2 + n(n − 1)(n − 2)x3 + · · · + xn ≥ 1 + nx
2
6
6. The simplest route is to expand
"
#
2
b
4ac − b2
bx
b2
b2
c
2
a x+
+
= a x +
+ 2+ − 2
2a
4a2
a
4a
a 4a
= ax2 + bx + c.
If b2 ≤ 4ac then it is clear that
if f (x) ≥ 0 for all
f (x) ≥ 0. Conversely
b
2
x, then in particular f − 2a ≥ 0 and so 4ac − b ≥ 0 as required.
7.
f (x) =
n
X
(ai x + bi )2
i=1
=
n
X
!
a2i
2
x +2
i=1
n
X
i=1
!
ai b i
x+
n
X
b2i .
i=1
The result now follows by applying the result
P6b where we
Pn of2 Problem
takeP
a, b and c (respectively) to be a = i=1 ai , b = 2 i=1 ai bi and
c = ni=1 b2i .
2