Download MATH1025 ANSWERS TO TUTORIAL EXERCISES V 1. If a b has a

MATH1025
ANSWERS TO TUTORIAL EXERCISES V
a
1. If has a terminating decimal expansion, then it can be written as a
b
fraction whose denominator is a power of 10. When we cancel this to lowest
terms, the denominator is a factor of a power of 10, so cannot be divisible by
primes other than 2 or 5. Conversely, if b is not divisible by primes other than
2 and 5, it has the form 2r 5s for some r, s (possibly 0). Then a multiple of b is
a
a power of 10, so by multiplying top and bottom of by this multiple, we get
b
c
an expression for it in the form k , and this has a decimal expansion which
10
terminates after k decimal places.
2. First suppose that b has the stated form. Then some multiple of b is 99
c
a
1
. The expansion of 99
times a power of 10, so for some c and k, =
k
b
99.10
c
recurs in blocks of length 2 (it’s 0·01010101 . . .), and hence so does that of 99
(more formally, if we write c = 99q + 10a1 + a0 where 0 ≤ r < 99 and a0 , a1
c
c
lie between 0 and 9 then 99
= q·a1 a0 a1 a0 a1 a0 . . .). The expansion of 99.10
k is
obtained by shifting, so also recurs in blocks of length at most 2. Conversely, if
the expansion eventually recurs in blocks of length at most 2, then a multiple
by a power of 10 recurs immediately after the decimal point in such blocks.
But as we have seen the decimal part of such an expansion is just equal to a
fraction with a denominator of 99, so on cancelling to lowest terms we get an
expression as claimed.
3. An integer has a cube root if and only if its negative does, so we can
concentrate on the case of positive numbers which are at least 2. Let us write
a = pα1 1 pα2 2 . . . pαk k where these are powers of distinct primes. If each αi is a
multiple of 3, we can let αi = 3γi , and then pγ11 pγ22 . . . pγkk is an integer cube
root. Conversely, if a has pγ11 pγ22 . . . pγkk as an integer cube root then we find
that αi = 3γi (by uniqueness of the expression), and so each power is divisible
by 3.
(i) Now let a = pα1 1 pα2 2 . . . pαk k and b = q1β1 q2β2 . . . qlβk . Since a, b ∈ Z have no
factor in common, pi 6= qj , so the expression for ab is pα1 1 pα2 2 . . . pαk k q1β1 q2β2 . . . qlβk .
Hence ab has an integer cube root if and only if all αi and βj are divisible by
3, which holds if and only if a and b both have integer cube roots.
a
(ii) If a and b both have integer cube roots then an integer cube root of
b
a
is obtained as the quotient of their cube roots. Conversely, if has an integer
b
cube root, then by using the expressions for its numerator and denominator as
products of prime powers, we find that all the powers occurring are multiples
of 3, and so a and b both have integer cube roots.
4. We write a = 7q + r where 0 ≤ r ≤ 6. Then a2 = 49q 2 + 14qr + r2 , so
as a2 is divisible by 7, so is r2 . But the squares of 0, 1, 2, 3, 4, 5, 6 are 0, 1,
4, 9, 16, 25, 36 respectively, and of these only 0 is divisible by 7. Hence r = 0
which implies that a is divisible by 7.
√
√
Using this, now suppose for a contradiction that 7 is rational, 7 = ab in
lowest terms, where a and b are positive. Thus a2 = 7b2 . Hence a2 is divisible
by 7. By what we’ve just done, a is divisible by 7. Write a = 7c. Then
7b2 = 49c2 , so b2 = 7c2 . Hence b2 is divisible by 7. Hence b is divisible by 7,
which gives a contradiction (to ab in lowest terms).
√
5. Suppose for a contradiction that 7 is rational, and write it as ab where
a and b are positive integers. This means that a2 = 7b2 . Letting p1 p2 . . . pk
and q1 q2 . . . ql be the prime factorizations of a and b (where the primes listed
need not be distinct), we find that p21 p22 . . . p2k = 7q12 q22 . . . ql2 . We have a prime
factorization on the left with an even number of occurrences of primes, and on
the right with an odd number, which is contrary to the fundamental theorem
of arithmetic.
6. We cannot use the ‘classical’ proof by contradiction because the step
from a2 divisible by 12 to a divisible by 12 is false (for instance, consider a = 6).
We can use the alternative proof from the fundamental theorem of arithmetic,
since 12 has an odd number of prime factors
√ (counting 2 twice), though note
a modification does). In
that this wouldn’t apply as it stands to
√
√ 6 (though
12
=
2
3.
We
can show by the usual
this case it’s easier
to
argue
like
this:
√
√
methods that 3 is irrational, and hence so is 12.
2