Download Solutions to problem sheet 1.

MH, 14 Oct 2007
M1F Foundations of Analysis—Problem Sheet 1 solutions
√
√
(1∗ ) a. Suppose for a contradiction that 6 − 2 ≤ 1. For any positive numbers
x and y, if x ≤ y then x.x ≤ x.y ≤ y.y, that is x2 ≤ y 2 . Hence
√
√
√ √
( 6 − 2)2 = 6 − 2 2 6 + 2 ≤ 12 = 1,
√
and therefore 7 ≤ 2 12. Squaring both sides gives us 49 ≤ 48 which is a contradiction.
b. Suppose for a contradiction there exists an integer n such that n2 is even, but
n is odd. But the square of any odd number is also odd (Liebeck, Example 1.2).
Hence n2 is odd, which is a contradiction.
c. n = m3 − m = m(m2 − 1) = m(m − 1)(m + 1). Hence n is the product of
three successive integers. But any three successive integers contains both an even
number and a multiple of 3. Hence n has factors of both 2 and of 3 and hence is a
multiple of 6.
(2∗ ). Suppose for a contradiction that there is a positive real number x0 that is
less than every positive number. Then in particular, since x0 is a positive number,
we must have x0 < x0 , which is a contradiction.
(3∗ ) a. This is true. If r and s are any two rational numbers then r = m/n and
s = p/q for some integers m, n, p and q with n and q non-zero. Then
m0
mp
= 0,
nq
n
0
0
with m = mp and n = nq both√integers with n0 non-zero. Hence rs is rational.
b. This is false. Let x = y = 2. Then xy = x2 = 2, which is rational.
c. This is true. Suppose for a contradiction that x is a non-zero rational, y is
irrational and z = xy is rational. Since x is a non-zero rational, then so is 1/x.
Hence y = z.(1/x) is the product of two rational numbers and hence by a result
proved in the lectures is rational, which is a contradiction.
rs =
(4∗ ) a. is true. x ∈
/ S ⇔ x ≤ 0. x ≤ 0 implies x53 ≤ 0. y ∈ S ⇔ y > 0, which
53
implies y > 0. Hence x53 ≤ 0 < y 53 as claimed.
√
√
b. is true. Given
√ x > 0, define y = x/(2 2). Then since x > 0 and 2 2 > 2, we
have 0 < y = x/(2 2) < x/2 as required.
√
√
c. is false. A counterexample is x = 1. Since 3 > 1 any y satisfying y < 1 − 3
is negative and so does y ∈
/ S.
(5) Suppose for a contradiction that all prime factors of 8881 are greater than
89. Let p1 ≤ p2 ≤ . . . ≤ pn be all the √
prime factors of 8881 in non-decreasing
order. Then p1 must be no
larger
than
8881. In particular, since 902 = 8100
√
2
and 95 = 9025 then 90 < 8881 < 95 and hence p1 ∈ S = {90, 91, 92, 93, 94, 95}.
But there are no prime numbers in S since 90, 92, 94 are all even and 91 = 7 ∗ 13,
93 = 31 ∗ 3 and 95 = 19 ∗ 5. Hence we have a contradiction.
(6) This is an exercise to get you to think clearly about the meaning of notation,
as well as deductive reasoning. We have two rules valid for all n ≥ 1, namely:
(i) an+1 > an and (ii) aan = 3n.
(a) Rule (ii) tells us that aa1 = 3. In particular, this means that 3 is one of
the numbers in our sequence. Since the sequence is strictly increasing (rule (i)),
it follows that the sequence must begin 1, 2, 3, . . . or 1, 3, . . . or 2, 3, . . . . But if
a1 = 1, then rule (ii) would say 1 = a1 = aa1 = 3, which is absurd. Thus the only
possibility is that a1 = 2, and hence a2 = 3.
1
(b) We can now use rule (ii) to deduce two more numbers in the sequence:
a3 = aa2 = 3 × 2 = 6, a6 = aa3 = 3 × 3 = 9.
And knowing that the sequence has to be increasing, the only possible way of
filling-in the blanks in 2, 3, 6, , , 9, . . . is if a4 = 7 and a5 = 8.
Using rule (ii) again we get a7 = aa4 = 12 and a8 = aa5 = 15 and a9 = aa6 = 18.
(c) One can continue to argue like this indefinitely, using rule (ii) to deduce some
terms further down the sequence and then finding that there is exactly one way to
fill-in the gaps (as we did to find a4 and a5 ). Notice that at no point do we have
to guess that we recognise the sequence as something familiar — we prove what
must come next. Nevertheless, you could try to deduce the general pattern. Or you
could try to proceed quickly to the point of the sequence that interests you. For
example, we can see that a12 = 21, a21 = 36, a36 = 63, a63 = 108, a108 = 189 and
that a11 = 20, a20 = 33, a33 = 60, a60 = 99, a99 = 180. From which it follows
that between a99 = 180 and a108 = 189, the terms of the sequence only increase by
1 with each step. In particular a100 = 181.