MTH 447 Homework #3 Jake Smith

HW Part 1

Problem 1: Proposition. Whenever x is odd, x2 + 7x + 4 is even.

Proof. Suppose that we have some odd integer x. Then, by definition we can represent

x by the expression 2a+ 1, where a is some integer. Next we substitute 2a+ 1 into the

expression x2 + 7x + 4 in place of x.

(2a + 1)2 + 7(2a + 1) + 4 =

4a2 + 4a + 1 + 14a + 7 + 4 =

4a2 + 18a + 8 =

2(2a2 + 9a + 4) =

Thus, by definition of an even integer, we can see that whenever x is odd that x2+7x+4

is even.

Problem 2: Proposition. The Product of two rational numbers is rational.

Proof. Suppose that we have to rational numbers a and b. Then, by definition of a

rational number, a and b can expressed by a = pq, q 6= 0 and b = j

k, k 6= 0 where p, q, j, k

are integers. Now, taking the product of these two we find pq = pjqk

. Since we know

that the product of two non-zero integers is an integer, it is clear that pq is composed

of an integer divided by an integer. Therefore, by definition of a rational number pq is

rational.

Problem 3: Proposition. Let d,m and n be positive integers. Prove that if d|m and m|n, thend|n.

Proof. Suppose the we have integers d,m and n such that d|m and m|n. So, bydefinition of a divisor, we know that m = dk where k is some integer and n = ml

where l is some integer. Since we know that m = dk we can substitute that into our

expression for n. So n = ml = (dk)l = d(kl). Thus we see that d multiplied by the

product of k and l equals n. Therefore, d divides n.

Problem 4: Proposition. For positive integers d and n, if d|n and n is even, then 2d|n.

Proof. Notice that if n = 6 that 2|6. However, 2(2) = 4 and 4 - 6. Therefore, it is nottrue if d|n and n is even, then 2d|n.

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HW Part 2

Problem 1: Proposition. If n is an integer, then n is even if and only if 5n + 3 is odd

Proof. Suppose n is an even integer. Then, by definition of an even number we can

express n by n = 2a where a is some integer. Now, substituting n with 2a in 5n + 3

yields the expression 5(2a) + 3. This expression can be rewritten thusly, 2(5a+ 1) + 1.

Therefore, if n is even, 5n + 3 is odd.

Now, suppose that 5n + 3 is odd. Then, by definition we can write 5n + 3 = 2k + 1,

where k is some integer. Then, through simple algebraic manipulation we can write

the equality as 5n = 2(k 1), which implies that 5n is even. Since 5n is even, and 5is an odd integer, n must be an even integer to make 5n even. Therefore, n is an even

integer.

Problem 2: Proposition. If the product of two integers is even, then at least one of the integers

must be even.

Proof. Suppose we have two integer p and q and that for the sake of contradiction that

they are both odd. Then, by definition of an odd integer p = 2a + 1 where a is some

integer and q = 2b+ 1 where b is some integer. Taking the product of the two integers

we find,

pq =

(2a + 1)(2b + 1) =

4ab + 2a + 2b + 1 =

2(2ab + a + b) + 1

Clearly, this is a contradiction since the product of the two integers is, by definition,

an odd integer. Therefore, at least one of the integers must be even.

Problem 3: Proposition. If n is a positive integer, then 4 - n2 2.

Proof. Suppose that we have a positive integer n.

Problem 4: Return to the Island of Knights and Knaves.

Proof. Suppose that Abe is a knight, and is therefore telling the truth. Then, by our

hypothesis, Abe is telling the truth when he says that Debbie is a knave. Since Debbie

is a knave, she is lying when she says that one of either Chris or Beatrice is a knight.

Thus, Chris and Beatrice are both knaves and , therefore, liars. Since both Chris and

Beatrice stated that Abe is a knave and they are both liars, Abe must be a knight.

Therefore, Abe is the only knight, while Beatrice, Chris and Debbie are all knaves.

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