Solutions to Problem set #2
Math 313
1. Let m be a natural number. I will prove by induction on a natural number n that mn
is a natural number. For the base case n = 1, note that mn = m · 1 = m, which is
a natural number by hypothesis. For the induction step, suppose that n is a natural
number and that mn is a natural number; we must prove that m(n + 1) is a natural
number. We have m(n + 1) = mn + m, which is a sum of two natural numbers, and
is therefore a natural number by the result proved in class.
2. Fix an arbitrary natural number m. I will prove by induction on a natural number n
that one of the following conditions holds: (i) n − m is a natural number, or (ii) m − n
is a natural number, or (iii) m = n.
For the base case n = 1, there are two subcases depending on whether m is 1 or not.
In the subcase m = 1, (iii) holds. In the subcase m 6= 1, we can use the fact pointed
out in class that every natural number not equal to 1 has the form k + 1 for some
natural number k. This means that m − n = m − 1 = k, a natural number, and so
(ii) holds.
For the induction step, suppose that n is a natural number and that one of the
alternatives (i), (ii) or (iii) holds; we must prove that one of these alternatives still
holds when n is replaced by n + 1. If (i) holds, i.e. n − m is a natural number, then
(n + 1) − m = (n − m) + 1, and so (n + 1) − m is also a natural number, i.e. (i) still
holds when n is replaced by n + 1. If (iii) holds, i.e. m = n, then (n + 1) − m = 1,
a natural number, so that (i) n + 1. Finally, suppose that (ii) holds, i.e. m − n is a
natural number. Here there are two further subcases. If the natural number m − n is
not 1, then by the fact quoted above we have m − n = k + 1 for some natural number
k; hence m − (n + 1) = k, a natural number, so that (ii) still holds when n is replaced
by n + 1. Finally, if m − n = 1 then m = n + 1, i.e. (iii) holds when n is replaced by
n + 1.
3. (a) In class I proved that the sum of two natural numbers is a natural number.
Hence if x and y are both natural numbers then x + y is a natural number, and in
particular an integer. If x = 0 then x + y = y, which is an integer by hypothesis;
similarly, if y = 0 then x + y = x, which is an integer by hypothesis.
Now consider the case where x = m is a natural number and y = −n for some
natural number n. Since m and n are both natural numbers, one of the three
alternatives of Problem #2 holds. If (i) holds, i.e. n − m is a natural number,
then x + y = m − n = −(n − m) is an integer by definition. If (ii) holds, i.e. m − n
is a natural number, then x + y = m − n is a natural number, and in particular an
integer. If (iii) holds, i.e. m = n, then x + y = m − n = 0, which is by definition
an integer.
In the case where x is minus a natural number and y is a natural number, we are
in the case just done when we interchange x and y; thus y + x is an integer in
this case, and by commutativity of addition this means x + y is an integer.
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This leaves the case where x = −m and y = −n for some natural numbers m and
n. In this case, x + y = −(m + n). Since m and n are natural numbers, m + n is
a natural number by the result proved in class, and so −(m + n) is an integer by
definition.
(b) If x or y is 0 then xy = 0, an integer. If x and y are both natural numbers, then
xy is a natural number by Problem #1 above, hence an integer. If x = m is a
natural number and y = −n for some natural number n, then xy = −(mn), where
mn is a natural number by Problem #1. If x = m for some natural number m,
and y = n is a natural number, we are in the case just done when we interchange
x and y; thus yx is an integer in this case, and by commutativity of addition this
means xy is an integer. Finally, if x = −m and y = −n for some natural numbers
m and n, then xy = mn, which by Problem #1 above is a natural number and
hence an integer.
4. (a) Suppose that x = a/b and y = c/d, where a, b, c and d are integers, and b and
d are non-zero. By the laws of algebra that were deduced in class from the field
axioms, we have
c
ad
bc
ad + bc
a
+
=
.
x+y = + =
b
d
bd
bd
bd
By Problem #3, ad + bc and bd are integers, and since b and d are non-zero we
have bd 6= 0. This shows that x + y is rational.
Similarly,
a c
ac
· = ,
b d
bd
where ac and bd are integers by Problem #3, and bd 6= 0 since b and d are
non-zero. This shows that xy is rational.
xy =
(b) Again by the laws of algebra, if x = a/b, where a and b are integers and b 6= 0,
we have −x = (−a)/b, where −a is an integer because a is an integer. (This is
immediate from the definition.) Thus −x is rational. If x 6= 0 then a 6= 0, and
x−1 = b/a. Thus x−1 is rational.
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