# 18.100B Problem Set 1

**Problems.**

1) (10 pts) Prove that there is no rational number whose square is 12.

2) (10 pts) Let S be a non-empty subset of the real
numbers , bounded above. Show that if

u = sup S, then for every natural number n , the number u − 1/n is not an upper
bound of S,

but the number u + 1/n is an upper bound of S.

3) (10 pts) Show that if A and B are bounded subsets of R,
then A∪B is a bounded subset of

R. Show that

supA ∪ B = max{sup A, supB}.

4) (20 pts) Fix b > 1.

a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

Hence it makes sense to define
(How could it have failed to make sense?)

b) Proce that b^{r+s} = b^{r}b^{s} if r, s are rational.

c) If x is real, define B (x) to be the set of all numbers b^{t}, where t is
rational and t ≤ x.

Prove that

b^{r} = supB (r)

when r is rational . Hence it makes sense to define

b^{x} := supB (x)

for every real x .

d) Prove that b^{x+y} = b^{x}b^{y} for all real x and y.

5) (10 pts) Prove that no order can be defined in the
complex field that turns it into an ordered

field.

(Hint: −1 is a square .)

6) (10 pts) Suppose z = a + bi, w = c + di. Define

z < w if a < c or a = c, b < d

Prove that this turns the set of all complex numbers into
an ordered set. (This is known as a

dictionary order , or lexicographic order.) Does this ordered set have the
least -upper-bound

property ?

7) (10 pts) Prove that

|x + y|^{2} + |x − y|^{2} = 2|x|^{2} + 2|y|^{2}

if x ∈ R^{k} and y ∈
R^{k}. Interpret this geometrically, as a statement about parallelograms.

**Extra problems:**

1) (Another argument showing that
)

Show that, if n^{2} = 2m^{2}, then

(2m − n)^{2} = 2(n − m)^{2}.

Deduce that, if n and m are strictly positive integers
with n^{2} = 2m^{2}, we can find strictly

positive integers n', m' with (n')^{2} = 2(m')^{2} and n' < n. Conclude that the
equation

n^{2} = 2m^{2} has no non-zero integer solutions.

2) Show that the square root of an integer is either an
integer or irrational.

(Hint: Every integer has a unique ( up to order ) factorization into a product of
prime numbers,

you can use this to show that if n is an integer and a prime p divides n ^{2}, then
p divides n.)

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