English | Español

Try our Free Online Math Solver!

Online Math Solver

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


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 AB 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 br+s = brbs 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
br = supB (r)

when r is rational . Hence it makes sense to define
bx := supB (x)

for every real x .
d) Prove that bx+y = bxby 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 Rk and y Rk. Interpret this geometrically, as a statement about parallelograms.

Extra problems:
1) (Another argument showing that )
Show that, if n2 = 2m2, then
(2m − n)2 = 2(n − m)2.

Deduce that, if n and m are strictly positive integers with n2 = 2m2, we can find strictly
positive integers n', m' with (n')2 = 2(m')2 and n' < n. Conclude that the equation
n2 = 2m2 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.)

Prev Next