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These are the course notes for MIT's calculus class, 18.014.

Beginning at the bottom of page 4, the professor launches into a long proof that the set of all integers $\mathbb Z$ is closed under addition.

I don't see why I can't demonstrate it by reasoning the following way. All integers in $\mathbb Z$ are either elements of $P$ (the set of positive integers), the negative of $P$'s elements, or $0$. $P$ is defined as the set of elements common to all inductive sets. So $P$ therefore contains $1$ and $x+1$, for any $x$. Any numbers $a,b$ in $P$ is an integer by definition, and is composed of adding $1+1$ multiple times. Therefore, adding or subtracting $a$ and $b$ is really adding or subtracting $1$ multiple times.

Which means that the set of all integers is closed under addition and subtraction, since you can only get integers.

So why is the level of rigor in the lecture notes necessary?

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You could enter that in the World's Longest Run-on Sentence competition. Or, you could edit it into something comprehensible. –  Gerry Myerson Aug 19 '12 at 9:44
    
It's readable now! –  terriblestudent Aug 19 '12 at 9:57
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@GerryMyerson. First version of question was very clear. Sarcastical alusion to competition is impolite, specially for new user. Why you think you may speak like that to people here? –  evgeniamerkulova Aug 19 '12 at 10:33
    
In fact, I don't understand why do you need to spend this amount of time and effort to prove such a thing. There are more interesting things at higher levels to work on them. –  a.r. Aug 19 '12 at 10:37
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@evg, I figure anyone who takes the handle "terriblestudent" probably has a good enough sense of humor to take my comment in the light-hearted way it was intended. –  Gerry Myerson Aug 19 '12 at 12:38

2 Answers 2

When you write "is composed of adding $1+1$ multiple times" you seem to be assuming that you already know how the integers work, because you're implying an argument about counting the number of terms in an expression that evaluates to your result.

There's at least two problems with that. First, you're not really proving anything from your axioms themselves -- at best you're proving that your axioms describe the same intuitive notion about integers that you're already using to reason about them, but that doesn't really satisfy the purpose of making clear which fundamental assumptions you're really basing your work on.

Second you're assuming that every element of $P$ can be written as $1+1+\cdots+1$, but you haven't any proof of that, from the principles you're starting with. And I doubt that any such proof can exist at the point in the development you're at -- your induction axiom is a bit vauge about which kind of properties it applies to, but I don't think it is intended to allow speaking about things such as "$1+1+\cdots+1$" before you have already gotten a solid foundation of the integers.


(By the way, I find the direction of the notes a bit strange -- they assume you already know what the real numbers and sets of them are and from there attempt to define the integers in an axiomatic foundational way. That is not a very orthodox way to introduce rigor. It may be that there is no really satisfactory way to motivate the choices made along the way from a principled mathematical point of view -- you may just have to roll with the flow.)_

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I think you're implicitly assuming the principle of induction is equivalent to the well-ordering principle here. Unless you've already proved this result, you can't appeal to that sort of reasoning.

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If a set doesn't extend down to $-\infty$, isn't it reasonable to assert that it has a smallest element? –  terriblestudent Aug 19 '12 at 9:57
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You mean, like the set of rational numbers $\gt7$? –  Did Aug 19 '12 at 10:01
    
Yes. But I'm also confused elsewhere. Where did I assume the well-ordering principle? Also, my only knowledge of the well-ordering principle is from having just read Wikipedia. –  terriblestudent Aug 19 '12 at 10:07
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There you go, @did just gave a perfect example of the subtlety involved in your assumption. –  anegligibleperson Aug 19 '12 at 10:09

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