Download M1F Foundations of Analysis Problem Sheet 2

M1F Foundations of Analysis
Problem Sheet 2
1. Prove or disprove the following statements:
(a) the sum of two irrational numbers is always irrational
False.
√
√
2 and − 2 are irrational numbers, but their sum is rational.
(b) the sum of a rational number and an irrational number is always irrational.
True. Suppose not. Then there exist integers p, q 6= 0, p0 , q 0 6= 0 and an irrational number i
such that p/q + i = p0 /q 0 .
Rearranging gives i = (p0 q − pq 0 )/qq 0 , which is rational. Contradiction.
(c) if n and k are positive integers, then nk − n is always divisible by k
False. 24 − 2 = 14 is not divisible by 4.
(d) ∀n ∈ N ∃a, b ∈ N such that n = a2 + b2 .
To prove this is false I must show that ∃n ∈ N such that ∀a, b ∈ N, n 6= a2 + b2 .
Consider n = 3. Assume for a contradiction that n = a2 + b2 .
Then a2 ≤ 3 so a < 2; similarly b < 2. Therefore the only possibilities for a and b are 0 and
1 (or just 1, depending on your view of N). Therefore the only possibilities for a2 + b2 are
0, 1, 2. But n = 3. Contradiction.
2. What is the smallest number in the set {x: x > 1} ? Justify your answer.
There is no smallest number. The number 1 is not in the set, and any number x > 1 cannot be
the smallest number because there is always a smaller number in the set, such as (1 + x)/2.
3. Write down a careful proof that for any two positive numbers x, y, their mean
(or average, or “arithmetic mean”) 12 (x + y) is greater than or equal to their
√
“geometric mean” xy:
x+y √
≥ xy.
2
When are they equal ?
BE VERY CAREFUL WITH THIS PROOF: ARE YOUR IMPLICATIONS IN THE CORRECT
DIRECTION ?!
√
Suppose for a contradiction that it is not true, i.e. x + y < 2 xy.
Squaring both sides, x2 + 2xy + y 2 < 4xy ⇒ x2 − 2xy + y 2 < 0 ⇒ (x − y)2 < 0. Contradiction.
Equality holds iff it holds at every step of the above calculation, iff (x − y)2 = 0. That is, if and
only if x = y.
4. Prove by induction that 7n − 3n is divisible by 4 for all n ≥ 1.
(Can you also see a direct, one-line proof ?)
True for n = 1: 7 − 3 = 4.
Assume true for n = k ≥ 1. So 7k − 3k is divisible by 4.
Then for n = k + 1, 7k+1 − 3k+1 = 7.7k − 3.3k = 7.(7k − 3k ) + 4.3k is also divisible by 4.
h
Direct proof: notice that xn − an is divisible by x − a. (In fact xn − an = (x − a)(xn−1 + axn−2 +
i
. . . + an−2 x + an−1 ).) Now put x = 7, a = 3.
5. * For which n ∈ N is 2n + 3n < 4n ?
Experimentation seems to show that it is true for n ≥ 2. It is clearly false for n = 0, 1.
So prove by induction that it is true for n ≥ 2. True for n = 2: 22 + 32 = 13 < 16 = 42 .
Suppose true for n = k. Then for n = k + 1, 2n + 3n = 2k+1 + 3k+1 = 2.2k + 3.3k ≤ 3(2k + 3k ), which
by the induction assumption is in turn < 3.4k < 4.4k = 4k+1 .
6. Suppose that an integer n is the sum of two squares (the squares being the
numbers 0, 1, 4, 9, 16, 25, ....). Show that 2n is also. (There is a small trick involved in
solving this; don’t worry if you can’t spot it, just look at the answer next week.)
Write n = a2 + b2 for integers a, b.
Then (a + b)2 + (a − b)2 = a2 + 2ab + b2 + a2 − 2ab + b2 = 2(a2 + b2 ) = 2n.
7. Prove that for every positive integer n, the number
Suppose
Write as
√
√
2+
√
n is irrational.
√
2 + n = r is rational.
√
√
√
n = r − 2 and square to give n = r 2 − 2r 2 + 2.
So either r = 0 (impossible; r ≥
√
2) or
√
2=
r 2 +2−n
.
2r
But this is rational, a contradiction.
You should prepare starred questions * to discuss with your personal tutor.