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) = ab2df((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 x3 = 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 x3 = 6. Then I can write x = p/q, where p, q ∈ Z satisfy q > 0
and
gcd(p, q) = 1. Since x3 = 6, I infer p3 = 6q3. Since 2|6 and 6|p3, it follows
that 2|p3. Since 2 is prime, it further follows that 2|p. Now I write p = 2k
for some k ∈ Z. The equation p3 = 6q3 becomes 23k3 = 6q3, or a little more
simply 22k3 = 3q3. In particular, 2|3q3. Since 2 is prime and does not divide
3, I infer that 2|q3 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 x3 = 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 xn = 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 xn = 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
xn = m, I obtain
an = mpnkn.
In particular, p|an 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.