The work here is actual assigned homework from Dr. Mary Flagg’s Number Theory, Math 4383 – Section 19842, going on now at the University of Houston this Spring 2012 semester. This is from homework assignment 2. The problems are taken from Elementary Number Theory, Seventh Edition, by David. M. Burton.
However, I will be going beyond much of this homework by adding additional material as it strikes me as being relevant to the topic at hand.
Basic Divisibility Theory of the Integers
2.2 – 5. For , prove that is an integer.
Proof. By the division algorithm, if , we may write
where . Therefore, since r may only assume integer values, we have that may only assume any one of the following six forms, corresponding to each value of the remainder so that n must be in one of the following forms:
, , , , ,
Therefore, we need only substitute each of these in our expression and simplify to see what results. So, proceeding on a case by case basis, we have, for the first case:
for we get
which is an integer, as there are no fractions left on the RHS. Now, in the second case:
for we get
which is also clearly an integer. And, in the third case:
for we get
which is also clearly an integer. And, in the fourth case:
for we get
which is also clearly an integer. And, in the fifth case:
for we get
which is also clearly an integer. And, in the sixth case:
for we get
and this too is an integer. Therefore, for all values of , we have established that
is an integer for all . QED
Note. Of course, this expression is also the sum of the squares of the first n integers, or
for all and so, because the square of an integer is an integer, so would be a sum of squares of integers, and so this may be developed out into another proof, as a corollary, by obtaining and proving the summation formula itself it would follow that it is an integer.
2.2 – 11. If n is an odd integer, show that is of the form 16k.
Proof. Assume that n has the form so that n is odd. Then, by direct substitution and upon simplification, we have
Therefore, is of the form 16K for all odd positive integers. QED
2.3 – 3. Prove or disprove: if , then either or .
Proof. By counterexample, we have
but false and is false.
So, without loss of generality, we may say that,
if and but , then it is true that:
or more fully, that and we see that and are both false. QED.
2.3 – 4. For , use mathematical induction to establish the following divisibility statement that
Proof. Using the PMI, we have for ,
and therefore it is true that
when n=1. Assume that for some positive integer k that
is true and note that this also means that:
for some multiple q of 15. But the LHS factors into:
which implies that:
and now this implies that, either:
and likewise for 5, that either:
one of which must be true and for each of the integers 3 and 5. Now, for (k+1) we have that:
where q is the quotient and r must equal 0 if 15 is to divide exactly into . Now, by factoring, we have that:
so that we have also:
but each of these factors is simply the same as the assumed case n=k when n=k+1. And it is already true, by assumption, that
Therefore, by the PMI, we have that,
, . QED
2.3 – 13 Given two integers a and b, prove the following: there exists integers x and y for which if and only if
Proof. This is almost the converse of theorem 2.4 on page 23, namely that, if a and b are both integers not equal to zero, then a and b are relatively prime if and only if there exist integers x and y such that . From this we have that, if , then:
but from corollary 1 to the theorem 2.4, both on page 23, we have that if then and this implies that:
but multiplying both sides of this by d and setting d=c yields:
but this means that but c|a and c|b and therefore because . QED
2.3 – 13(b) If there exist integers x and y for which , then .
Proof. This is theorem 2.4 where the terms are inverted. We have, by theorem 2.4, that a and b are relatively prime if and only if there exists integers x and you such that . Now, suppose that for some choice of x and y, such that . Because and , by theorm 2.2 we get that or but this last condition forces d to be 1. Therefore, must be true. QED
2.3 – 20(e). Confirm the following properties of the greatest common divisor. If and and , then .
Proof. We are given that a and b are relatively prime, which means that:
for some linear combination of a and b their exist such integers x and y that this is true. Now, if and means that and for some q, q’, r, r’. Therefore, we must have that d divides either a or c, given that , and likewise, d must divide either b or c, given that .
Assume, in the first case, that d divides a but not c. And assume that, in the second case, that d divides b but not c also. Then we would have and or subtracting these from the other set, that:
but a and b are relatively prime implies, if d dives both a and b, then:
but since it is also true that and that each implies that and that the respectably, which is a contradiction. Therefore, d must divide c if and . QED
2.3 – 21.
(a) Prove that if , then .
Proof. First off, if , then there exists a q and r such that:
where and so, we are being asked to prove that:
Now, from first principles, because , we must have that:
or else there could be no division by d into n in integer terms, and so therefore, we also have:
for some positive integer because . Note that this implies that . Now, subtracting one from both sides yields:
And, because implies that , and this implies that , upon division of the RHS by the LHS, yields:
, then , and so, if the , then
for some . Now, without loss of generally, taking and
substituting, we have:
but this implies that
or dividing both sides by b
which implies that
and likewise argument if we go substituting but this is just and therefore, if , then . QED
2.4 – 9. Prove that the greatest common divisor of two positive numbers a and b divides their least common multiple, especially that .
Proof. By merging the definitions of the gcd and lcm of two integers, a and b, we have that the gcd of a and b is the positive integer d satisfying the following:
(b) if and , then
and the lcm of two positive integers a and b is the positive integer m satisfying:
(b) If and , with , then
but and , therefore and because and . Again, we have that and , therefore and because and . Therefore, we see that the gcd does indeed divide the lcm, and we have established that
Saturday 28 January 2012
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