1.4 Exercise 5

I had often wondered how to go about proving closure of addition and multiplication of the integers.  After this problem I wondered no more!  It’s pretty neat that we can show (by using induction) that by adding any two integers or multiplying them you always get another integer.

“Prove the following properties of and :

(a) .  [Hint: Show that given , the set and is inductive.]

(b) .

(c) Show that .  [Hint: Let and ; show that is inductive.]

(d) and . [Hint: Prove it first for .]

(e) .”

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 35.)

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SOLUTION

(a)

Using the hint, we proceed by

• Showing .  Since , and is inductive, it follows that .   Therefore .
• Supposing .  This of course means .
• Showing .  This would mean .  Note that by commutativity  and associativity (of the reals, of the integers), we can state .  But then we know that by the first point, and that by the second point.  So we conclude that indeed .

Thus is inductive.  In particular, then, .  It follows that if we pick , with by hypothesis, then we are guaranteed .

(b)

This is proved the same way as before, with a few minor modifications.  Again suppose .  Let and .  Again we want to show that is inductive.

• , since (by identity axiom of the reals).
• Assume . This means .
• , since (by the distributive property axiom of the reals), with by the first point and by the second point.  Finally, by what we proved in part (a).

We conclude is inductive, and contains .  As before, with , pick , and we are guaranteed the product .

(c)

Again following the hint, we show:

• .  This is because is true.
• Suppose , which means that .
• We show that .  This would mean that , or, simplifying, that , which is true because is a positive integer by definition.

Thus, is inductive and .  Picking , we see that .

(d)

Taking a hint from previous hints (pun intended), assume .  Next define and . We want to show that is inductive.

• .  First of all, we want to show that .  Certainly, by part (a).  If , then .  If , then with , so by part (c), and (!). Secondly, we want to show that . Then, by part (c). If , .  Finally, if , then can be rewritten as with , and by part (a).  Then certainly .  Thus .
• Suppose that , which means and that .
• We show that .  First, means, by rearranging (associativity, commutativity) .  But we know by the first point, and so by the second.  We conclude that indeed .  Next we show .  By the distributive property, .  Then by associativity, .  We know by the first point, and by the second point. We conclude that indeed .

Thus, is inductive, and importantly, . For any integer , picking an integer guarantees that and .  If , then simply is true by hypothesis.  This shows quite thoroughly the closure of addition of the integers!

(e)

Again taking a hint from previous hints, assume .  Next define and .  We want to show is inductive.

• . It seems clear that , since simplification means , which is true by hypothesis.
• We assume that , which means .
• We show that . This means we will show that .  By the distributive property, .  Now, we know by the first point, and by the second.  Thus by part (d), and indeed .

Thus is inductive and .  Having , pick and we are guaranteed that .  Next, if , then .  If , make the negative sign explicit, so that and by the properties of the reals this becomes with and , which again guarantees the product to be in .  We have just proved closure of multiplication of the integers!