Download 1. Consider the function f : N × N → N × N defined by f

1. Consider the function f : N × N → N × N defined by
f (m, n) = (2m + n + 1, n).
(i) Is f an injection? Supply a proof or a counterexample with explanation.
(ii) Is f a surjection? Supply a proof or a counterexample with explanation.
Solution. (i) yes. Let (a, b), (c, d) be elements of N × N such that
f (a, b) = f (c, d). Then 2a + b + 1 = 2c + d + 1 and b = d. Substituting b = d into the previous equation yields a = c, so (a, b) = (c, d) and f is
an injection.
(ii) no. (0, 0) is not in the image of f since 2m + n + 1 6= 0 for all (m, n)
in N × N.
2. (i) Let p be a prime number. Show that
[x2 ≡ y 2 mod p] implies [(x ≡ y mod p) or (x ≡ −y mod p)].
(ii) Provide an example of a natural number n and integers x and y such
that [x2 ≡ y 2 mod n] and [x 6≡ y mod n] and [x 6≡ −y mod n].
Solution. (i) x2 ≡ y 2 mod p implies that p | (x − y)(x + y), and since
p is prime this means that p | (x − y) or p | (x + y), ie (x ≡ y mod p) or
(x ≡ −y mod p).
(ii) [42 ≡ 22 mod 12] and [4 6≡ 2 mod 12] and [4 6≡ −2 mod 12].
3. Let x and y be integers satisfying x4 + x2 − 8y = 0. Prove that 4 | x.
(Hint: mod 8)
Solution. x4 + x2 = 8y, so x2 (x2 + 1) ≡ 0 mod 8. x = 0 and x = 4 work,
but one can verify that 12 (12 + 1), 22 (22 + 1),32 (32 + 1),52 (52 + 1),62 (62 + 1),
and 72 (72 + 1) are not congruent to 0 mod 8. Thus x = 8k or x = 4 + 8k
for some integer k and in either case we see that 4 | x.
4. Use the Euclidean Algorithm to compute gcd (471, 297) and express the
greatest common divisor as an integer combination of 471 and 297. Show
your work (you may use the extended Euclidean algorithm).
Solution. 471 = 297 · 1 + 174, 297 = 174 · 1 + 123, 174 = 123 · 1 + 51,
123 = 51 · 2 + 21, 51 = 21 · 2 + 9, 21 = 9 · 2 + 3, 9 = 3 · 3 + 0.
Back substituting non-zero remainders yields 3 = 46 · 297 + (−29) · 471
5. Let S be the set of all ordered pairs of positive real numbers (x, y) and
let T be the set of all ordered pairs of positive real numbers (a, b) such that
a2 > b. Show that f : S → T defined by f (x, y) = (x + y, x2 ) is a bijection.
Solution. Let (x, y) and (s, t) be in S and assume that f (x, y) = f (s, t).
Then (x + y, x2 ) = (s + t, s2 ) so x2 = s2 and x + y = s + t. Since x and s are
positive, the first equation implies that x = s, and substituting this into the
second equation implies that y = t. So√(x, y) =
√ (s, t) and f is an injection.
√
Now
let
(a,
b)
be
in
T.
Consider
(
b,
a
−
b).
Since
b
>
0
and
a
>
b,
√
√
√
√
( b, a − b) is in S. Moreover, f ( b, a − b) = (a, b), and thus f is a
surjection.
Since f is an injection and a surjection, f is a bijection.
6. Show that if a and b are relatively prime integers, then the greatest
common divisor of 2a2 and b2 must be 1 or 2.
Solution. Let d be the greatest common divisor of 2a2 and b2 . Note that
no prime p larger than 2 can divide d, since p would divide a2 and b2 and
hence divide both a and b, in conflict with the relative primality of a and b.
Also, no multiple of 4 can divide d since this implies that a2 and b2 are even,
which implies that a and b are even, in conflict with the relative primality
of a and b. Thus, the only possible factor of d which is larger than 1 is 2,
and hence d must be 1 or 2.
7. (i) Find the remainder of 9123 when divided by 61.
(ii) Find the remainder of 9123 when divided by 77.
Solution. (i) By FLT, 960 is congruent to 1 mod 61. Thus, 9123 ≡
≡ 58 mod 61.
(ii) By Euler’s theorem, 960 is congruent to 1 mod 77. Thus, 9123 ≡
(960 )2 93 ≡ 36 mod 77.
(960 )2 93
8. Verify that the function f : R3 → R3 defined by
f (x, y, z) = (x + y, 2y + z, z − x)
is a bijection.
Solution. Let (x, y, z) and (r, s, t) be in R3 and assume that f (x, y, z) =
f (r, s, t). Then x + y = r + s, 2y + z = 2s + t and z − x = t − r, so,
y = −(x + y) + (2y + z) − (z − x) = −(r + s) + (2s + t) − (t − r) = s.
Substituting y = s into x + y = r + s implies that x = r, and substituting
x = r into z − x = t − r implies that z = t. Thus, (x, y, z) = (r, s, t) and f
is an injection.
Let (a, b, c) be in R3 and consider (2a − b + c, −a + b − c, 2a − b + 2c) in
R3 . f (2a − b + c, −a + b − c, 2a − b + 2c) = (−a + b − c + 2a − b + c, 2(−a + b −
c) + 2a − b + 2c, 2a − b + 2c − (2a − b + c)) = (a, b, c). Thus, f is a surjection.
Since f is an injection and a surjection, f is a bijection.