# Math Homework #7 Solutions

**Problem 1.** Prove each of the following (using
properties of integers and the relevant

definitions from class concerning rational numbers):

(a) Addition of rational numbers is well-defined.

(b) Addition of rational numbers is commutative.

(c) The distributive law holds for rational numbers.

Solution. Note that in what follows, whenever I write a rational number
or

an ordered pair (a, b), I assume automatically that a ∈ Z and b ∈
.

(a) Suppose that (a, b) ~ (a', b') and (c, d) ~ (c', d'). That is, ab' = a'b and

cd' = c'd. Then according to our definitions, we have

(a, b) + (c, d) = (ad + bc, bd) and (a', b') + (c', d') = (a'd' + b'c', b'd').

Comparing the results, we find

(ad + bc)(b'd') - (a'd' + b'c')(bd)

= (adb'd' - a'd'bd) + (bcb'd' - b'c'bd)

= dd'(ab' - a'b) + bb'(cd' - c'd)

= dd' · 0 + bb' · 0 = 0,

since ab' = a'b and cd' = c'd. I conclude that

(a, b) + (c, d) ~ (a', b') + (c', d').

Hence addition respects the equivalence relation ~, and addition of rational

numbers is well-defined.

(b) Given rational numbers and
, I have

This proves that addition of rational numbers is commutative.

(c) Given rational numbers , and
, I have on the one hand that

and on the other hand that

To verify that the results are the same, I cross-multiply and check

(acf + aed)(bdbf) - (acbf + aebd)(bdf) = ab^{2}df((cf + ed) - (cf + ed))
= 0.

So multiplication of rational numbers is distributive over addition.

**Problem 2.** The wrong way to add fractions. Given
rational numbers and
, suppose we

define a 'new' kind of sum

(a) While it's appealingly simple, this definition of 'addition' has the slight
drawback

that it gives answers that are inconsistent with our expectations (e.g.

?). However, all expectations aside, this
definition has a worse flaw: it isn't even

consistent with itself. Give an example to illustrate.

(b) On the other hand, the above idea for adding rational numbers isn't entirely
without

merit (this is a major understatement—if you ask me about it, I'll explain).
Show

that if , then
. (I should point out that this is another
way to see

that Q has the density property.)

Solution.

(a) Note, for instance,

whereas

However, . Hence ⊕ can't be a genuine
operation on rational numbers,

because it depends on how the numbers are expressed, rather than

just on the numbers themselves.

(b) First I show that

a(b + d) - (a + c)b = ad - bc < 0|

because .

Similarly, to see that , I cross-multiply and
check

(a + c)d - c(b + d) = ad - bc < 0

**Problem 3. **Show that is irrational.
That is, there is no x ∈ Q such that x^{3} = 6. Of

course, you want to imitate the proof that
is irrational. But be careful: 6 is not prime.

Solution.

Proof. Suppose (in order to get a contradiction) that there exists x ∈ Q such

that x^{3} = 6. Then I can write x = p/q, where p, q ∈ Z satisfy q > 0
and

gcd(p, q) = 1. Since x^{3} = 6, I infer p^{3} = 6q^{3}. Since 2|6 and 6|p^{3}, it follows

that 2|p^{3}. Since 2 is prime, it further follows that 2|p. Now I write p = 2k

for some k ∈ Z. The equation p^{3} = 6q^{3} becomes 2^{3}k^{3} = 6q^{3}, or a little more

simply 2^{2}k^{3} = 3q^{3}. In particular, 2|3q^{3}. Since 2 is prime and does not divide

3, I infer that 2|q^{3} and therefore 2|q. But since I
already showed that 2|p, this

contradicts the assumption that p and q are relatively prime. I conclude that

there is no x ∈ Q satisfying x^{3} = 6.

**Problem 4.** Let n ≥ 2 and m ≥ 0 be integers. Show that
is rational if and only if

is an integer. More precisely, show that

if x ∈ Q satisfies x^{n} = m, then x is actually an integer.

Hint: There are a couple of different ways to do this, each resembling (but not
perfectly)

the proof that is irrational. Note that a rational number x = a/b given in
lowest terms is

an integer if and only if b ≥ 2. It might help to think first about the case
where b is prime.

Solution.

Proof. Suppose that x^{n} = m for some x ∈ Q. I can write x =
for some

integers a, b such that gcd(a, b) = 1. Suppose to get a contradiction that x

is not an integer. Then b ≥ 2 and must have a prime factor: b = pk for

some prime number p and integer k. Substituting x = a/pk into the equation

x^{n} = m, I obtain

a^{n} = mp^{n}k^{n}.

In particular, p|a^{n} which, since p is prime, means that p|a. This contradicts

the fact that a and b have no common factors larger than 1. Hence x must be

an integer after all.