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I have just recently started to work with Calculus by Spivak and I am just wondering some things about the first chapter. ( I am doing this as a method to review my calculus which I have done but only using texts such as Stewart).

It is about the basic principals of numbers and such. However, I do not have an answer booklet, so I am unsure if what I am doing is correct or not. (Which is also kinda hard, should I try to find a book that gives answers)?

But overall , my question is in regard to all these types of problems. I am looking for someone to motivate why and how important it is to work through all of these problems in the first chapter of spivak or similar ,

And For an example ,

One question is,

Prove $$\frac{a}{b}=\frac{ac}{bc}$$ if $b,c \neq 0$

So to me , I just thought I will use what had been talked about in the chapter.

Ie, $$\frac{ac}{bc}=(ac)(bc)^{-1}=(ac)(b^{-1}c^{-1})=(ab^{-1})(cc^{-1})=(ab^{-1})$$

But I just don't know at all if that is what is expected, I am looking for any other resources to understand exactly what I am supposed to do. Further, it seems like it can't be correct ( or atleast not the way that was supposed to be done) because a few problems after this it asks to prove that $$(ab)^{-1}=a^{-1}b^{-1}$$

So can anyone help provide some insight/advice/ recommendations or anything?

I just want to actually make sure I am understanding what is correct and expected for these type of questions. Also, if anyone has any specific insight on how to do the proof without using the property that comes later, or even if possible someone who has Spivak to say a method using only what had been given.

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    $\begingroup$ It's not "trivial". You are asked to prove well-known properties of numbers starting from first principles (or axioms). I don't know how Spivak treats rational numbers, but equality of rational numbers is usually defined by $\frac{a}{b}=\frac{ac}{bc}\iff a(bc)=(ac)b$. This is true in virtue of the the associativity and commutativity of $(\mathbb{Z},\cdot)$, which should be known by then. I believe your proof would be OK if you knew $(ab)^{-1}=a^{-1}b^{-1}$ (which isn't hard to verify). $\endgroup$
    – Guest
    Commented Jun 23, 2015 at 2:50
  • $\begingroup$ Yea I know it is not trivial , I am not trying to say that. I just meant it felt that way $\endgroup$
    – Quality
    Commented Jun 23, 2015 at 2:53
  • $\begingroup$ @Quality I did this problem the same way you did and also encountered having to justify $(bc)^{-1}=(b^{-1}c^{-1})$, which we prove like 3 exercises later. $\endgroup$
    – user265675
    Commented Jun 23, 2015 at 2:53
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    $\begingroup$ @Quality - In general, when you're talking about the fundamental properties of very "basic" objects, e.g. $\mathbb{N}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$, different authors will introduce them in different ways. It's not necessarily that they have vastly differing ideas of what a real number is, but rather that each has different ways of understanding the building blocks (though these differences will not make a significant difference). When posing questions like these, it's helpful to explain what the author gave you (e.g. a rational is an ordered pair $(m, n)$ of such-and-such form...). $\endgroup$
    – AJY
    Commented Jun 23, 2015 at 5:58

1 Answer 1

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You could do the following.

basic properties:

1)$\forall a \neq 0 : \exists a^{-1}$ such that $ a^{-1}\cdot a = 1$

2)$\forall c :\; 1\cdot c = c$

3)$\forall a,b \; a\cdot b = b\cdot a$

4)$ \forall a,b,c \; (a\cdot b) \cdot c = a\cdot (b \cdot c)$

Now use these properties:

$$\frac{a}{b} = (a \cdot b^{-1}) = (a \cdot b^{-1}) \cdot 1 =(a \cdot b^{-1})\cdot \big( b\cdot c) \cdot (b \cdot c)^{-1}\big) = (a \cdot b^{-1})\cdot \big( b\cdot (c \cdot (b \cdot c)^{-1})\big) =(a \cdot (b^{-1} \cdot b) )\cdot (c \cdot (b \cdot c)^{-1}) = (a \cdot (1))\cdot (c \cdot (b \cdot c)^{-1}) =((a \cdot 1)\cdot c) \cdot (b \cdot c)^{-1} \\=(a \cdot c) (b \cdot c)^{-1} = \frac{ac}{bc}$$

This is a bit long, but we are using the basic properties one at the time

Remark: $a,b,c$ are real numbers.It is also valid for $a,b,c$ complex numbers - more generally for $a,b,c$ in any Field (see https://en.wikipedia.org/wiki/Field_%28mathematics%29).

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