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# 18.100B Problem Set 1 Solutions

1) The proof is by contradiction. Assume ∃r ∈ Q such that r2 = 12. Then we may write r as
with a, b ∈ Z and we can assume that a and b have no common factors . Then

so 12b2 = a2.

Notice that 3 divides 12b2 and hence 3 divides a2. It follows that 3 has to divide a (one way
to see this: every integer can be written as either 3n,3n+1, or3n+2 for some integer n. If you
square these three choices, only the first one gives you a multiple of three.)

Let a =3k, for k ∈ Z. Then substitution yields 12b2 = (3k)2 =9k2, so dividing by 3 we have
4b2 =3k2, so 3 divides 4b2 and hence 3 divides b2. Just as for a, this implies that b has to divide
b. But then a and b share the common factor of 3, which contradicts our choice of representation
of r. So there is no rational number whose square is 12.

2) S ⊆ R, S ≠ Ø, and u = sup S. Given any n ∈ N, ∀s ∈ S, , so is an upper
bound for S. Assume is also an upper bound for S. Since , u would not be the
least upper bound for S, which is a contradiction. Therefore is not an upper bound for S.

3) Recall that a subset of the real numbers , A ⊆ R, is bounded if there are real numbers a and a'
such that

t ∈ A ⇒ a' ≤ t ≤ a.

Since A, B ⊆ R are bounded, they have upper bounds a and b respectively, and lower bounds a'
and b'. Let α = max (a, b) and β = min (a', b'). Clearly,

t ∈ A ⇒ β ≤ a' ≤ t ≤ a ≤ α
t ∈ B ⇒ β ≤ b' ≤ t ≤ b ≤ α,

hence any t ∈ A ∪ B satisfies β ≤ t ≤ α and A ∪ B is bounded.

Notice that, in particular, this shows that max{sup A, supB} is an upper bound for A∪B, so
we only have to show that it is the least upper bound. Suppose γ< max{sup A, supB}. Then
without loss of generality, γ< supA. By definition of supremum, γ is not an upper bound of A,
so ∃a ∈ A with γ<a. But a ∈ A ⇒ a ∈ A ∪ B, so γ is not an upper bound of A ∪ B. Therefore
max{sup A, supB} = supA ∪ B.

4) Start by noting that, if n, m ∈ N then from which it follows that for
n, m ∈ Z (why?). Similarly, you can show that for n, m ∈ Z. Recall that, if x> 0,
then is defined to be the unique positive real number such that .

a ) We have that m/n = p/q so mq = pn. Notice that and that
, which is also equal to . But we know that there is a unique real
number y satisfying hence the two numbers we started with have to be equal, i.e.,

Notice that if this equality didn't hold, then we could not make sense of the symbol b r for
r ∈ Q, because the value would change if we wrote the same number r in two different ways.

b) Let r, s ∈ Q with and . Since nq is an integer we know that

but and similarly . Since mq and np are
integers we can conclude

.
But there is a unique positive real number, y, such that , so we know that

c) Now with b> 1, given r, s ∈ Q, s ≤ r we want to show bs ≤ br. Let ,0 <n,0 ≤ m
since s ≤ r. Then , and it is easy to see that 1 ≤ bm, since 0 ≤ m and 1 <b.
Thus a positive power of b r-s is greater than or equal to 1, which implies 1 ≤ br-s. Multiplying
by bs gives , so bs ≤ br . Hence for any ,
so br is an upper bound for B(r). Since br ∈ B(r), br must be the least upper bound, so
br = supB(r).

d) So let x, y ∈ R. If r, s ∈ Q are such that r ≤ x, s ≤ y, then r + s ≤ x + y so ∈ B(x + y)
and . Keeping s fixed, notice that for any r ≤ x we have

thus is an upper bound for B (x) which implies . We rearrange this to

and conclude that or .

Suppose the inequality is strict . Then ∃t ∈ Q, t<x + y, such that . We will find
r, s ∈ Q, with r ≤ x, s ≤ y and t<r+s<x+y. First, find N ∈ N so that N (x + y - t) > 1,
then find r ∈ Q so that and s ∈ Q such that (the existence
of N, r, s follow from the Archimedean property of R as shown in class). Now, notice that

hence we have t<r + s<x + y just like we wanted .

But now we have

which is a contradiction because, since r<x and s<y, we have br <bx and bs <by!

5) We know that in any ordered field , squares are greater than or equal to zero . Since i2 = -1, this
means that 0 ≤-1. Butthen1=0+1 ≤-1+1=0 ≤ 1 which implies 0 = 1, a contradiction!

6) I'll write for this relation on C to distinguish it from the normal order on R. To show that
is an order on C, we must show both transitivity and totality (or given x, y ∈ C, exactly one of
the following is true: x y, y x, or x = y). First for transitivity, let x, y, z ∈ C, x = a + bi,
y = c + di, z = e + fi such that x y z. Therefore a ≤ c ≤ e, so a ≤ e by the transitivity of
the order on R. If a<e, then x z, so we are done. If a = e, then a = c = e so we have from
the definition of that b<d<f, so once again by the transitivity of the order on R, b<f.
Now a = e and b<f ⇒ x z, so we have shown transitivity.

Now to show totality. Consider x, y ∈ C, x = a + bi, y = c + di. Without loss of generality,
let a ≤ c. Suppose a = c. Then b<d ⇔ x y, b>d ⇔ y x, and b = d ⇔ x = y, so by the
totality of the order on R, we have the totality of on C in the case of a = c. Suppose instead
that a<c. Then we know x y, and it is not the case that y x or x = y, so we have totality
in this case as well. Thus we have proven that is an order on C.

This order does not have the least-upper-bound property. Consider the set of complex numbers
with real part less than or equal to zero:

S = {a + bi : a ≤ 0,b ∈ R}.

S is bounded above, for instance by the number 1, but it is not possible for any number z = a+bi
to be the supremum of S. If a ≤ 0, then a + bi a +(b + 1)i ∈ S, so a + bi is not an upper
bound for S. If a> 0, then a +(b - 1)i a + bi, and a +(b - 1)i is also an upper bound for S,
so a + bi is not the least upper bound. Therefore S has no least upper bound, even though it is
bounded above.

The geometric interpretation comes from looking at the parallelogram whose vertices are the
points 0, x, x + y and y. Then the equation states that the sum of the squares of the lengths
of the two diagonals (the vectors x + y and x - y) is the same as the sum of the squares of the
lengths of the four sides.

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