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62
Solutions to Exercises
8. (a) First we show that
f
is onetoone. Suppose then that
f(xt}
=
f(X2).
If
XI and
X2
are both
negative, then 21xII
=
21x21. Since Ixl
=
x
for
x
<
0, we have 2xI
=
2X2
and hence,
XI
=
X2.
If
XI and
X2
are both
~
0, we have 2xI + 1
=
2X2
+ 1, so 2xI
=
2X2
and again,
XI
=
X2.
Finally, we note that it is impossible for
f(xt}
=
f(X2)
with XI
<
0 and
X2
~
0 (or
vice versa) since in this case, one of
f(xt}, f(X2)
is an even integer while the other is odd.
Next we show that
f
is onto. Suppose that
n
E N.
If
n
=
2k
is even, then
n
=
21 
kl
=
f(
k)
since
k
<
0, while if
n
=
2k
+
1 is odd, then
n
=
f(k)
since
k
~
O. Since
f
is onetoone and
onto, it has an inverse.
(b) Let
fI(2586)
=
a.
We must solve the equation 2586
=
f(a).
Since 2586 is even, we see that
2586
=
f(
2586/2)
=
f( 1293).
9.
(a)
[BB]
maternal grandmother
(d) motherinlaw
(g)
you
(b)
paternal grandmother
(e) father
.
(h)
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 Summer '10
 any
 Graph Theory

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