Mathematics 220 Homework #4 Solution

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Problem 1. Prove the following statement:

For every a Z such that 5 does not divide a, there exists b Z such that ab 1 mod 5.

Problem 2. We recall that given a, b Z such that ab = 0, we define the gcd of a and b to be the greatest integer that divides both a and b. We denote it by gcd(a, b).

1. Let a, b Z such that ab = 0. We suppose that there exists u, v Z such that

1 = au + bv.

Prove that gcd(a, b) = 1.

2. Write the negation of the following statement. Then decide if it is true or false and justify your answer.

For every a Z, b Z, and d Z such that d > 0,

(if there exists u, v Z such that d = au + bv, then gcd(a, b) = d).

Problem 3. Read the Division Algorithm theorem carefully in your book (Fact 1.5 page 30 in the third edition). What is the division of

1. 167 by 8?

2. 167 by 8?

Problem 4. Prove the following statement:

for every n Z such that n 0, we have 5|33n+1 + 2n+1 .

Problem 5. For each of the following statements:

Negate the statement,

Decide if the statement (prior to negation) is true or false and justify your answer.

1. a Z, if 6|a and 8|a then 48|a.

2. x R, y R, if xy 0 then x + y 0.

3. For every house h in Vancouver, there is a positive integer c such that (if the number of cats in h is > c, then there is no mouse in the house).

Problem 6. Explain what is wrong in the following proof and write a correct version of it.

For any real number b such that 1 < b < 5, prove that b3 6b2 + 5b < 0.

Proof. Suppose that 1 < b < 5.

b3 6b2 + 5b = b(b2 6b + 5) < 0 so, since b > 0 we have b2 6b + 5 < 0. But we notice that b2 6b + 5 = (b 1)(b 5). Since 1 < b < 5, we know that b 1 > 0 and b 5 < 0 so indeed b2 6b + 5 = (b 1)(b 5) < 0.

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