Download Math 58 - Number Theory

SOLUTIONS TO
Homework 2 (due September 12):
(Most of these exercises are from the text, App. A, 1.3, or 1.5.)
Math 58 - Number Theory
September 12, 2006
Prove rigorously using only the axioms in Appendix A:
1. If a and b are integers, then (–a)(–b) = ab.
First: If a and b are any integers, then a(-b) = -(ab).
Proof: “-(ab)” is, by definition, the unique number x that solves
ab + x = 0.
Well, a(-b) does solve that equation, because
ab + a(-b) = ba + (-b)a
(two applications of commutativity,
needed because we only have the
left-hand form of distributivity)
= (b + (-b)) a
(distributivity)
=0a
(def. of –b, and axiom on negatives)
= 0.
(by text example,
or imbed that proof here)
That shows that a(-b) = -(ab). //
Now, the problem as given.
Proof: From the above (with –a in place of a):
(-a)(-b) = -((-a)b)
= -(b(-a))
(commutativity)
= -(-(ba))
(from the above, with b,a in place of
a,b)
= ba
(since –(–x) = x…did we prove that?)
= ab. //
2. If a 
then a2 ≥ 0.
Proof: By the ordering axiom, either a is positive, or a=0, or –a is positive.
If a is positive, then a2 = aa is the product of two positive numbers, so
it is positive, so a2 > 0.
If a = 0, well, then a2 ≥ 0 by definition of ≥.
If –a is positive, then a2 = aa = (-a)(-a) (by problem 1)
which is the product of two positive numbers. So again, a2 > 0. //
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3. If a < b and c < 0 then ac > bc.
Proof: a<b means that b-a is positive, and c<0 means 0-c = -c is positive. So,
(b-a)(-c) must be positive.
By distributivity (I’ll apply it on the right this time),
b(-c) – a(-c) is positive
so -(bc) +(ac) is positive
so (ac) + (-(bc)) is positive
so (ac) – (bc) is positive
So ac > bc.
Prove by induction (or any variant on induction):
4. If n ≥ 4 then 2n < n!.
Proof: BWOC, let n be the smallest integer such that n≥4 but 2n is not < n!.
If n can’t be 4, because 24 = 16 < 24 = 4!.
So the statement must be true for n-1 in place of n:
2n-1 < (n-1)!
So
2n = 2 (2n-1) < 2 (n-1)! < n (n-1)! = n!. //
Prove by any method:
(Extreme rigor not required.)
5. If x and y are integers (different from each other) then x – y | x2 – y2.
(You can skip this one if you are satisfied with your proof of #7.)
Proof: All we need to do is show that
(x-y) c = (x2 – y2)
for some c. But this is true when c = x + y, because of the
factorisation
x2-y2 = (x-y)(x+y). //
6. If x and y are integers (different from each other) and n is any positive
integer, then x – y | xn – yn.
Proof: Now all we need is the factorization
xn – yn = (x – y ) ( xn-1 + xn-2y + xn-3y2 + … + yn-1).
7. If a, b, c, d 
//
, a and c are nonzero, and a | b and c | d, then ac | bd.
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Proof: If a | b then there is some x such that ax = b; and similarly,
there is some y
such that cy = d. So
(ac) (xy) = bd,
which is enough to force ac | bd. //
Prove OR disprove:
(If the statement is false, a single example is enough to disprove it. If any of these
is true, induction might be helpful in proving it.)
8.
If a is any integer, then a2 – a is even.
It’s true. Proof by factoring:
a2 – a = a ( a-1 )
and either a or a-1 must be even, so their product is even.//
9.
If a is any integer, then a3 – a is a multiple of 3.
True again: Now a3 – a = a ( a-1) (a+1), and the factors are three
consecutive integers, so one of them must be a multiple of 3, so
a3-a is also a multiple of 3.
10. If a is any integer, then a4 – a is a multiple of 4.
This isn’t always true. It’s false when a=2 or when a=6, for example.
11. If a is any integer, then a5 – a is a multiple of 5.
Two ways. Both use the expansion
(x+y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5.
Note that all terms of the expansion are multiples of 5 except for
possibly the first and last. If x is itself a multiple of 5, then
all but the last term are guaranteed to be multiples of 5.
First: Write a = 5q + r, where r must be 0, 1, 2, 3, or 4. Now:
a5 – a = (5q+r)5 – (5q+r)
= (  lots of multiples of 5 ) + ( r5 – r )
Now, r must be 0, 1, 2, 3, or 4, so (r5 – r) must be
0, 0, 30, 240, or 1020 --- in every case, another multiple of 5.
So, a5 – a is a sum and difference of multiples of 5, so is itself
a multiple of 5.
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Second: At least for positive a, try inducting on a. The statement is clearly
true when a is 0 or 1. Assume that it’s true for (a-1).
Now, a5-a = ( (a-1) + 1 )5 – a
= (a-1)5 + (multiples of 5) + 1 – a
= (multiples of 5) + ( (a-1)5 – (a-1) ).
By induction, the last expression is a multiple of 5, so a5-a is also
a multiple of 5.
(For negative a, note that a5-a = - [ (-a)5-(-a) ], and the expression in
square brackets is a multiple of 5, so its negative is, also.)
(end)
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