Download Math 261 Spring 2014 Final Exam May 5, 2014 1. Give a statement

Math 261
Spring 2014
Final Exam
May 5, 2014
1. Give a statement or the definition for ONE of the following in each category. Circle the letter next to
the one you want graded. For an extra good final impression, you can give the definitions/statements
for non-circled answers too! (Watch your time though.)
(A) The Well-Ordering Axiom.
The Well-Ordering Axiom says that every non-empty set of natural numbers has a least element.
(B) The Division Algorithm
The Division Algorithm says for natural numbers m and n there exist unique q and r such that
m = qn + r
and 0 ≤ r < n.
State either one below (or both, for added impression). Circle the letter of the one you want graded.
(A) The definition of a prime number.
A prime number is a natural number p that cannot be written as p = a · b for values a and b
that are less than p.
(B) The definition of a composite number.
A composite number is a natural number n that can be written as n = a · b for values a and b
that are less than n.
State either one below (or both, for added impression). Circle the letter of the one you want graded.
(A) The FTOA (Existence) Include all hypotheses and conclusions.
The FTOA (Existence) states that every natural number greater than 1 is either prime or can
be written as a product of positive powers of primes. In symbols, n > 1 has a prime factorization
r
n = pr11 · · · pj j ,
where pi is prime and ri > 0 for all i.
(B) The FTOA (Uniqueness) Include all hypotheses and conclusions.
The FTOA (Uniqueness) states that the prime powers in the prime factorization for a natural
number are unique and the primes used in the prime factorization are unique up to the order they
are written.
State either one below (or both, for added impression). Circle the letter of the one you want graded.
(A) The definition of a Complete Residue System Modulo n.
A Complete Residue System Modulo n is a set of a values {a1 , a2 , . . . , ak } such that any
integer x is congruent modulo n to one of the values in the set. In symbols, given an integer x we
must have
x ≡ ai (mod n)
for some ai in our set.
(B) The definition of the Canonical Complete Residue System Modulo n.
The Canonical Complete Residue System Modulo n is the set
{0, 1, 2, . . . , n − 1}.
State either one below (or both, for added impression). Circle the letter of the one you want graded.
(A) The definition of ordn (a).
For natural numbers a and n with (a, n) = 1, ordn (a) is the smallest natural number k such that
ak ≡ 1(mod n).
(B) The definition of the Euler-φ function φ(n).
The Euler-φ function φ(n) counts the number of values less than or equal to n > 0 that are
relatively prime to n.
State either one below (or both, for added impression). Circle the letter of the one you want graded.
(A) State Fermat’s Little Theorem (any version). Include all hypotheses and conclusions.
Fermat’s Little Theorem (Version I) says for p a prime and (a, p) = 1, we have
ap−1 ≡ 1(mod p).
Equivalently, Fermat’s Little Theorem (Version II) says for p a prime and any integer a, we
have ap ≡ a(mod p).
(B) State Euler’s Theorem. Include all hypotheses and conclusions.
Euler’s Theorem says for n > 0 and (a, n) = 1, we have aφ(n) ≡ 1(mod n).
2. Make a calculation (with some supporting justification) for ONE of the following in each category.
Circle the letter next to the one you want graded. For an extra good final impression, you can make
the other calculations too! (Watch your time though.)
(A) Find x and y to write 96x + 21y = (96, 21).
Begin by using the Euclidean Algorithm to find (96, 21):
96 = 4 · 21 + 12
21 = 1 · 12 + 9
12 = 1 · 9 + 3
9=3·3+0
So (96, 21) = 3. Then using the 3rd equation we know (96, 21) = 3 = 12 − 1 · 9.
Rewriting the 2nd equation gives 9 = 21 − 1 · 12 and this combines with the previous equation to
give
(96, 21) = 12 − 1 · (21 − 1 · 12) = 2 · 12 − 1 · 21.
Finally, rewriting the 1st equation gives 12 = 96 − 4 · 21. Combining this with the previous
equation
(96, 21) = 2 · 12 − 1 · 21 = 2 · (96 − 4 · 21) − 1 · 21 = 2 · 96 − 9 · 21.
So x = 2 and y = −9.
(B) Find the number of zeroes at the end of the normal (base ten) representation of 24!. By the
FTOA, every factor in 24! can be written as a product of primes. In particular, we are interested
in the number of times the primes 2 and 5 appear since a zero at the end of a number corresponds
to a factor of 10 = 2 · 5.
Note that every other (even) factor in 24! contributes at least one power of 2, with 4 = 22 ,
8 = 23 , and 12 = 22 · 3, 16 = 24 , 20 = 22 · 5 and 24 = 22 · 32 contributing more. So there are
2 · 22 · 2 · 23 · 2 · 22 · 2 · 24 · 2 · 22 · 2 · 22 = 220 powers of 2 in the prime factorization of 24!.
Only every fifth factor in 24! contributes a power of 5, and these come from 5, 10 = 5 · 2, 15 = 5 · 3,
and 20 = 5 · 4. So there are 5 · 5 · 5 · 5 = 54 powers of 5 in the prime factorization of 24!.
Each pair of 2 and 5 (multiplied together) contributes one more zero at the end of 15!. Since there
are only four such pairs (because there are only four factors of 5), there must be FOUR zeroes
at the end of 24!.
Question 2 continued . . . Calculate either one below (or both, for added impression). Circle the letter
of the one you want graded.
(A) Find the CCRS value corresponding to 267 (mod 31).
By Fermat’s Little Theorem (Version I), since (2, 31) = 1 we have
230 ≡ 1(mod 31).
Since 67 = 2 · 30 + 7 we must have
267 = 22·30+7 = (230 )2 · 27 ≡ 12 · 27 (mod 31).
It is easy to calculate 27 = 128. The correct CRS value a is such that 128 − a is divisible by 31.
Note that 4 · 31 = 124. So 128 − 4 = 124 = 4 · 31 is divisible by 31. Therefore
267 ≡ 4(mod 31).
We could also perform this calculation by noting that ord31 (2) = 5 since 25 ≡ 1(mod 31).
(B) Find any CRS value that is the inverse of 5(mod 57).
First note that 57 = 3 · 19 is a composite number. Since (5, 57) = 1, Euler’s Theorem says
5φ(57) ≡ 1(mod 57).
From our work in class, φ(57) = φ(3) · φ(19) (since 3 and 19 are relatively prime) and it follows
that φ(57) = 2 · 18 = 36. Then
536 ≡ 1(mod 57).
Then b = 535 is a CRS value that is the inverse since
5 · 535 = 536 ≡ 1(mod 57).
3. Let a, b, and n be integers with n > 0. The equation ax ≡ b(mod n) has a solution if and only if
(a, n)|b. You may cite only Chapter 1 results without proof.
Proof: We will break this proof into two parts corresponding to the “if and only if”.
Suppose ax ≡ b(mod n) has a solution. This means there exists a value for x such that n|(ax − b). By
definition of divides, there is an integer z such that nz = ax − b. Then ax − nz = b. Taking y = −z
gives integers x and y with
ax + ny = b.
Now by Chapter 1 results, this equation only has solutions if (a, n)|b.
Now suppose (a, n)|b with b = (a, n) · k for some integer k. By Chapter 1 results, ax̂ + nŷ = (a, n) for
some integers x̂ and ŷ. Then ax̂k + nŷk = (a, n) · k = b. Now nŷk = ax̂k − b or n|(ax̂k − b). Then by
definition of congruence ax̂k ≡ b(mod n) gives a solution to ax ≡ b(mod n) where x = x̂k.
4. Prove ONE of the following statements (or both, for added impression). Clearly indicate which proof
you are writing. If you write both, put one on the reverse side:
a2
(A) Show that there are no two natural numbers a and b such that 2 = 3. (This is a necessary step
b
√
to prove 3 is irrational, but you don’t have to prove this latter statement)
Proof: Assume for a contradiction that
a2
= 3. This implies that a2 = 3b2 .
b2
r
By FTOA, both a and b have prime factorizations, say a = pr11 · · · pj j and b = q1s1 · · · qkrk . Then
by algebra a2 and b2 have prime factorizations
2rj
1
a2 = p2r
1 · · · pj
and b2 = q12s1 · · · qk2rk .
Since each exponent of the primes in the factorization of a2 is multiplied by 2, every power must
be even. Similarly, each exponent of the primes in the factorization of b2 is multiplied by 2, every
power must be even. However, a2 = 3b2 implies that the power of 3 in the prime factorization
of a2 must be odd, since 3b2 gives an odd (even plus one) number of occurrences of 3. This is
contradicts that FTOA says prime factorizations are unique, but earlier we saw all powers in the
prime factorization of a2 were even. Therefore, we cannot have two natural numbers a and b such
a2
that 2 = 3.
b
a2
= 3 for some natural numbers a and b. Then consider the set
b2
√
√
S = {k 3| k and k 3 are natural numbers}.
√
What can you say about this set? Try to use this set to prove 3 is irrational.
(B) Assume
a2
= 3 for some natural numbers a and b and consider the set S.
b2
√
√
a √
a2
Since 2 = 3 we have = 3 and a = b 3. Since a is a natural number, b 3 must be a natural
b
b
number. This implies that S is a non-empty set of natural numbers. By WOA, S must have a
least element, call it `.
√
Consider
We can show 3 is irrational by demonstrating our assumptions lead to a contradiction.
√
our least element `. Since ` ∈ S, there must be some natural number k with ` = k 3.
√
√ √
It follows that ` · 3 = k · 3 · 3 = 3k is also a natural number.
√
√
Now check this out: ` 3 − ` = `( 3 − 1) = 3k − ` must also be a natural number.
√
√
√
√
Also ( 3 − 1) > ( 1 − 1) = 0 and 1 = ( 4 − 1) > ( 3 − 1).
√
√
√
√
But then `( 3 − 1) is a natural number less than `. Multiplying `( 3 − 1) by
√ 3 gives 3` − 3`
which, by supposition, is the natural number 3` − 3k. These imply that `( 3 − 1) is in S, but
this contradicts our choice of ` as the least element. Therefore, we cannot have natural numbers
a2
a and b with 2 = 3.
b
Proof: Assume
6
5. Prove ONE of the following statements (or both, for added impression). Clearly indicate which proof
you are writing. If you write both, put one on the reverse side:
(A) If p and q are distinct primes, then pq−1 + q p−1 = 1(mod p · q).
Proof: By Fermat’s Little Theorem (Version I) pq−1 ≡ 1(mod q) and q p−1 ≡ 1(mod p).
Note since p ≡ 0(mod p) and q ≡ 0(mod q) we also have pq−1 ≡ 0q−1 (mod p) ≡ 0(mod p) and
q p−1 ≡ 0p−1 (mod q) ≡ 0(mod q).
Putting together the pieces modulo p and q now gives:
pq−1 + q p−1 ≡ 0 + 1(mod p) ≡ 1(mod p)
pq−1 + q p−1 ≡ 1 + 0(mod q) ≡ 1(mod q)
Now by Theorem 4.21 above, we have pq−1 + q p−1 ≡ 1(mod p · q).
(B) If p is prime and a is an integer with 1 < a < p then p|(ap + (p − 1)!a).
Proof: We must show that p|(ap + (p − 1)!a) or that ap + (p − 1)!a = k · p for some k ∈ Z.
By Fermat’s Little theorem, we know that
ap ≡ a (mod p).
By Wilson’s Theorem,
(p − 1)! ≡ −1 (mod p),
and it follows (using a Ch. 1 result) that a(p − 1)! ≡ −a (mod p). Then
ap + (p − 1)!a ≡ a + (p − 1)!a (mod p)
≡ a + (−a) (mod p)
By FLT and Chapter 1 results
From what follows after Wilson’s Theorem
≡ 0 (mod p)
But this says that ap + (p − 1)!a − 0 = k · p for some k ∈ Z. So p|(ap + (p − 1)!a).
6. Show n5 − n is divisible by 5 for every natural number n by writing a proof by induction. If you cannot
prove it using induction, prove it another way (for reduced credit).
Proof: Begin by considering the base case when n = 1. Since 15 − 1 = 0 = 0 · 5, we see that 5|(12 − 1).
Assume, as our induction hypothesis that the result holds up to n. That is, for every value i less than
or equal to n we have 5|(i5 − i). In particular 5|(n5 − n) and n5 − n = 5j for some integer j.
We now need to consider n + 1.
Note that (n + 1)5 − (n + 1) can be written as
n5 + 5n4 + 10n3 + 10n2 + 5n + 1 − (n + 1) = n5 − n + 5(n4 + 2n3 + 2n2 + n).
By our induction hypothesis, 5|(n5 − n) and n5 − n = 5j. Then
n + 1)5 − (n + 1) = 5j + 5(n4 + 2n3 + 2n2 + n) = 5(j + n4 + 2n3 + 2n2 + n).
This shows that 5|((n + 1)5 − (n + 1)) and the desired conclusion follows from proof by induction.
Alternate, Reduced-Credit Proof: We know that p = 5 is a prime. Then by Fermat’s Little
Theorem (Version II)
a5 ≡ a(mod 5)
for all natural numbers a. By definition of congruence, 5|(a5 − a) for such values of a.
If instead we use Fermat’s Little Theorem (Version I) then for (a, 5) = 1 we have
a4 ≡ 1(mod 5).
Multiplying this equation by a now gives a5 ≡ a(mod 5) and the result follows as above.
For a not relatively prime to 5, we must have 5|a since 5 is a prime. Then a ≡ 0(mod 5) and
a5 ≡ 05 ≡ 0(mod 5). By Chapter 1 results,
a5 − a ≡ 0 − 0(mod 5) ≡ 0(mod 5).
Then by definition of congruence, this says that 5|(a5 − a − 0) or 5|(a5 − a).