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146 The Nuts and Bolts of Proof, Third Edition
x — y = 0, or x = y. The proof is not complete because we still have to
prove that this is the only possible conclusion. The second equaUty
can be rewritten as (x - y)^ = -1. Because (x-yf is always nonnegative,
this equality will never be true. Therefore, the product
(x-yf[{x-yf-\-i] is equal to zero only if (x-yf =0; that is,
when x = y.
Part 2. We have to prove that if x = y, then (x - yf -\-{x- yf = 0.
This is quite easy to do; indeed, in this case x — y = 0.
2. Part 1. These two sequences are equal if x" = y" for all n>2. Because
x^ = y^, we obtain (x — y)(x -f- y) = 0. Therefore, we have two possible
conclusions: either x = y or x = —y. We can only accept the conclusion
x = y. Indeed, if x = —y, then x^ = -y^. But x^ = y^ by hypothesis.
Part 2. The converse of this statement is trivial.
3. By definition, a divides b if the division of b by a yields a counting
number and zero remainder. Therefore, we can write b/a = q, with q
counting number. Similarly c/b = t, where t is a counting number, and
a/c = s, where s is a counting number. These three equalities can be
rewritten as b = aq, c = bt, a = cs. If we use all of them, we obtain
b = (cs)q = c(sq) = (bt)(sq) = b(tsq) (#). Therefore, b = b{tsq). Because
b ^0 (since 0 is not a counting number), we obtain 1 = tsq. As t, s, and
q are all counting numbers and are larger than or equal to 1, tsq can
equal 1 if and only if t = 1, s = 1, and ^ = 1. If we use this result in (#),
we obtain b = a c = b a = c. Therefore, a = b = c.
4. Let a, ft, and c be three counting numbers. Set d = GCD{ac, be) and
e = GCD(a, b). We want to prove that d — ce.
Part 1. d>ce. Because e = GCD(a, b), we can write a = ke and b = se
with k and s relatively prime. Multiplying both equalities by c, we
obtain ac = k{ce) and bc = s{ce). This proves that ce is a common
divisor of ac and be. But d is the greatest common divisor. Thus, d > ce.
Part 2. d< ce. Because e is the greatest common divisor of a and ft, we
can write a = ke and b = se, where k and s are relatively prime. So,
multiplying by c, we obtain ac = k{ce) and be = s{ce), where k and s are
relatively prime. Then all the common factors of ac and be are in ce.
Thus, ce is larger than any other common divisor. So d< ce. From the
two parts of this proof we can conclude that d = ce.
5. By hypothesis (a/b)^ = n, where n is an integer. Thus, a^ = b^n or
a^ =1 b{b^~^n). This means that b divides a^. Because a and b are
relatively prime, their greatest common divisor is 1. Therefore, b and
a^ cannot have any common factors other than 1. Thus, b = l.