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Description
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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