MATH 310C
SPRING 2007
SAMPLE EXAM II SOLUTIONS
1. (a) FALSE. Here is a counterexample: f(x) = xand g(x) = −xare both bijections. (Why?)
But h(x) is not. (Why not?)
(b) TRUE.
Proof: Suppose fand gare strictly increasing and that x1and x2are real numbers
such that x1< x2. Since fis strictly increasing, x1< x2implies that f(x1)< f (x2).
Similarly, g(x1)< g(x2). This means that f(x1) + g(x1)< f(x2) + g(x2) and by the
definition of h(x), h(x1)< h(x2). Thus, his strictly increasing.
2. (a) Since d|aand d|b, there exist integers kand k0such that a=dk and b=dk0. So, a
d=k
and b
d=k0, which are both integers.
(b) Let kand k0be as defined in part (a). Then gcd a
d,b
d= gcd(k, k0). Let n= gcd(k, k0).
Then n|kand n|k0. This means that there exist integers k1and k0
1such that k=nk1
and k0=nk0
1. But then a=dk =dnk1and b=dk0=dnk0
1, which means that dn is a
common divisor of aand b. If n > 1, then dn > d, which contradicts the fact that dis
the greatest common divisor of aand b. So, nmust be 1.
3. (a) Note that gis essentially the same function as f. The only difference is the target. So,
gis injective because fis injective and gis surjective by the definition of f(A). (The
target of gis precisely those elements that get “hit” by f(x) and therefore by g(x).)
(b) i. If |A|=nfor some n∈N, then there exists a bijection F:A→[n]. The
function gfrom part (a) is a bijection from Aonto f(A). As such, ghas an inverse,
g−1:f(A)→A. Since the composition of two bijections is a bijection, F◦g−1is a
bijection from f(A) onto [n]. Thus, |f(A)|=n=|A|.
ii. If Ais countable, then there exists a bijection from Aonto N. Again, using the
bijection gfrom part (a), g−1is a bijection from f(A) onto A. So, F◦g−1is a
bijection from f(A) onto N. Thus, f(A) is also countable.
iii. If f(A) is countable, then there exists a bijection Ffrom f(A) onto N. Again, with
g:A→f(A) the bijection from part (a), the function F◦gis a bijection from A
onto N. This means that Ais countable. This shows that, if f(A) countable, then
Ais countable. The converse is therefore true: if Ais uncountable, then f(A) is
uncountable.
4. (a) By the Division Algorithm, there exist unique kand rsuch that Q= 4k+rand
r= 0,1,2,or 3. Since Qis odd, rcannot be 0 or 2. So, rmust be 1 or 3. By the
uniqueness of r,Qis either of the form 4k+ 1 or 4k+ 3, but not both.
(b) Basis step: n= 1: a1∈S⇒a1∈SX
Induction step: Suppose that, if a1, a2, ..., ai∈S, then a1a2...ai∈Sand consider the
product a1a2...aiai+1, where a1, a2, ..., ai, ai+1 ∈S. By induction, a1a2...ai= 4k+ 1 for
some integer kand ai+1 = 4k0+ 1 for some integer k0. Then,
a1a2...aiai+1 = (4k+ 1)(4k0+ 1) = 16kk0+ 4k+ 4k0+ 1 = 4(4kk0+k+k0)+1∈S.
