Department of Mathematics

Colorado State University

Math 317: Homework 1

Due: Friday, February 1, 2019

Solutions to proof questions are not unique! There may be many different ways to prove any one statement. Remember this as you read solutions—these are only sample proofs!

  1. (1.1) Prove for all positive integers .

    Proof. By induction.

    Base case. holds because .

    Induction step. Assume holds. Consider the LHS of :

    where we have used the induction hypothesis. Factoring out from both terms yields that

    So holds, and by induction, all hold for all .

  2. (1.6) Prove is divisible by 7 when is a positive integer. (An integer is divisible by an integer if there exists an integer so that .)

    Proof. By induction.

    Base case. holds because which is trivially divisible by 7.

    Induction step. Assume holds; this means that we assume that there exists so that . Consider

    We have that so we get that

    where we have used the inductive hypothesis in the last step. So,

    as is an integer, we conclude that divides , hence that is true.

    So by induction, for all , is true.

  3. (4.3, partial) For each subset of below, determine both the supremum and the infimum, if they exist. If either doesn’t exist, say so. You do not need to give a rigorous proof of your answer.

    1. ; .

    2. ; . (For this question, it’s also okay if you say it DNE).

    3. ; . This set only has 3 elements.

    4. ; . Even though , it is still a subset of the real numbers, so it has an infimum and supremum (an we know what they are).

    5. ; . is certainly an upper bound. Proof that it is the least upper bound:

      Let . Then and so let be a natural number which exists by the Archimedean property. So then . So cannot be an upper bound. So is indeed the least upper bound.

  4. Don’t worry about writing out any formal proofs in this problem. Decide whether each of the following statements is true. If the statement is true, you don’t need to do anything more. If the statement is false, give a concrete example (that is, a counterexample) that shows the statement failing.

    1. For a nonempty, bounded set , .

      False. Let be a set of one element.

    2. If is rational and is irrational, then is irrational.

      True. Were for some integers then which would be a rational number (why?).

    3. If , is nonempty and is bounded, then .

      True. If is an upper bound for as (definition of subset) so . As is a least upper bound for , it must be at most as big as the upper bound .

    4. A finite, nonempty set always contains its supremum.

      True. Finite, nonempty sets have maximums, which agree with supremums when they exist.

  5. Let be a nonempty set of real numbers which is bounded below. Let be the set . Prove that

    (Note: This is the bulk of the proof of Corollary 4.5. You’re welcome to make use of all of the theorems and properties in Section 3, including Theorem 3.2. is an ordered field.)


    Let . For any element , we have that by definition of . By definition of supremum, we have then that . By theorem 3.2, multiplying both sides by yields that . As this holds for all , we conclude that is a lower bound for .

    Consider any lower bound of . This means that for all . Theorem 3.2 then says that for all , which means that for all by definition of . So is an upper bound for . As is the supremum of , it is lesser than any other upper bound so that . Theorem 3.2 then says that . This is true for any lower bound of , so is indeed the infimum of the set ; that is, .

    We conclude then that .