Are Assumptions in Spivak's Calculus a Common Theme Throughout the Book?

[Thinker301]

Hi everybody!

I have started working through Spivak's "Calculus". I was slightly upset with the first chapter. The book made a few assumptions. (The concept of a number and that if a=b then a+c=b+c, stuff like that) I am fine with it for now because I know where the gap in knowledge is but when I get further in the book, I won't be able to recognize things like that.

I was wondering if the book kept assuming things like that later in the book?

Thank you for your time!

 

[adjacent]

if a=b then a+c=b+c

It's not difficult.It's elementary algebra.
if a=b then you can replace one of those variables:
a+c=b+c - replace a with b because a=b.
Then you will get b+c=b+c
Two b's cancel out so you have c=c.

 

[Thinker301]

I apologize if this is rude, but I really can't tell, and I am sorry if it is. Are you making a joke?

 

[adjacent]

I apologize if this is rude, but I really can't tell, and I am sorry if it is. Are you making a joke?

Why do you think I am making a joke? I am sorry if I offended you by saying it's elementary algebra.

I said that because it's simple.
You were given that if $a=b$ then $a+c=b+c$
Here $c$ can be anything and if $a=b$ we can interchange $a$ and $b$.
So we can either write $b+c=b+c$ or $a+c=a+c$.
In both cases, the other variables cancel out and what's left is $c=c$ so it is correct to say that if $a=b$ then $a+c=b+c$.

Why were you uncomfortable with assumptions like that?

 

[Thinker301]

Ahhh ok.

Truly sorry about that. Really I am.

I misunderstood your proof.

I would actually rewrite it like this.

$a = b$ - Given
$a + c = a + c$ (I am not not quite sure what this is called)
$a + c = b + c$ because a = b.

I was a little confused when you talked about cancelling things out but this Theorem allows for one to cancel, and the conclusion wasn't $a + c = b + c$.

But anyways, I was able to catch that assumption when reading the book. I might not be able to catch a similar assumption later in the book. I am wondering if the rest of the book has these little assumptions.

EDIT : Also it assumes the definition of even or odd.

 

[462chevelle]

There is no need to rewrite anything. its just a fact. if c is equal to say.. 2.
that +2 on both sides doesn't even matter.since you could just subtract it from both sides and it states that a=b because c=c

 

[Thinker301]

The problem is you can't subtract yet. We are trying to prove the theorem which allows us to do that.

 

[Thinker301]

We can't add a number to both sides yet, that is the theorem we are trying to prove. Subtraction is just a consequence of being able to add, after we have the axiom of $a + (-a) = 0$

 

[462chevelle]

I wasn't saying to subtract it, just illustrating why the c is redundant.

 

[Thinker301]

I think there is some disconnect here. Our tangent was that given a=b, we had to prove a+c=b+c.

But anyways, does Spivak keep doing this throughout the book?

 

[micromass]

Ok, I see what you mean. It is true that Spivak doesn't prove it, and it cannot be proven. But you can take it as an axiom. Spivak should have stated it, but very few books do. If you ever study mathematical logic in depth then you might see a justification of this.

 

[micromass]

But anyways, does Spivak keep doing this throughout the book?

No, he doesn't. This is just one situation where he does it. After the first two chapters, such a thing shouldn't happen anymore.

 

[Thinker301]

I worked through the historical section of Apostol I think I saw it in there. I had to return Apostol back to the library so I can't check now.

Where is the fallacy in my revision of adjacent's proof?

Does
$a+c = a+c$ assume the conclusion?

And to the original point, thank you Micromass.

 

[micromass]

Ahhh ok.

Truly sorry about that. Really I am.

I misunderstood your proof.

I would actually rewrite it like this.

$a = b$ - Given
$a + c = a + c$ (I am not not quite sure what this is called)

OK. (the second line is an axiom of equality by the way.

$a + c = b + c$ because a = b.

I don't see how this follows from the previous two lines. You need an axiom or inference rule that does this.

I worked through the historical section of Apostol I think I saw it in there. I had to return Apostol back to the library so I can't check now.

Where is the fallacy in my revision of adjacent's proof?

And to the original point, thank you Micromass.

It is very rare that math books start by stating all the axioms. The only place you'll see this happening is in mathematical logic books. There, everything is built from the axioms rigorously. A site that does this is metamath: http://us.metamath.org/
But usually books assume some axioms. In this case, Spivak assumes the axiomatics of sets and of equality. This means specifically, that he assumes that if $A=B$ and that if $f$ is a function, then $f(A) = f(B)$. You can take this as an axiom. Furthermore, he assumes that addition is a function, thus $f(B) = B+C$ is a function. So it follows that if $A=B$, then $f(B) = f(A)$, and thus $B+C = A+C$.

 

[Thinker301]

I don't see how this follows from the previous two lines. You need an axiom or inference rule that does this.

Because the a=b, I replaced the second a with b.

I will certainly check the website out!

 

[adjacent]

You can take this as an axiom. Furthermore, he assumes that addition is a function, thus $f(B) = B+C$ is a function. So it follows that if $A=B$, then $f(B) = f(A)$, and thus $B+C = A+C$.

So you can take anything as a axiom? Like $1=3$?
You said that he assumes addition as a function. How does he assume a function then? Is everything ultimately reduced to mere axioms?

 

[micromass]

Because the a=b, I replaced the second a with b.

Sure, I see that. But who says you can do that?

 

[micromass]

So you can take anything as a axiom? Like $1=3$?

Yes, that would be a valid axiom, but one that does not reflect our world.

You said that he assumes addition as a function. How does he assume a function then?

By defintion, he considers a set $F$, equipped with a function $+:F\times F\rightarrow F$ that satisfies some axioms. This is the mathematical way of saying the things that Spivak means. There is also a function $\cdot: F\times F\rightarrow F$ and a relation $<$. All of these satsify some axioms.

Is everything ultimately reduced to mere axioms?

It should be, yes.

 

[Thinker301]

Hmmm, I see. I am new to this so this is probably very naive. If a=b, isn't a and b interchangeable?

EDIT : Nevermind, I looked at the page.

 

[micromass]

Hmmm, I see. I am new to this so this is probably very naive. If a=b, isn't a and b interchangeable?

Sure, it should be. But that's why you need axioms.

Here is a simple example, consider the operation $*$ on $\mathbb{Q}$ defined as follows:

$$\frac{a}{b}*\frac{c}{d} = \frac{a+c}{b+d}$$

This satisfies all of Spivak's axioms for $+$ (associative, zero element, inverses, commutative).

But then for example, we have $A = \frac{0}{1}$, $B=\frac{0}{2}$ and $C=\frac{1}{1}$. Certainly, $A=B$, but

$$A*C = \frac{0}{1}*\frac{1}{1} = \frac{0+1}{1+1} = \frac{1}{2}$$

and

$$B*C = \frac{0}{2}*\frac{1}{1} = \frac{0+1}{2+1} = \frac{1}{3}$$

So $A=B$, but $A*C\neq B*C$. So there are funny examples of "operations" that do not satisfy this. What goes wrong of course is that the function $f(A) = A*C$ is not a function, since $A=B$ does not imply here that $f(A) = f(B)$.

So Spivak should have added an extra axiom that $+$ is a function. Together with the axiom of equality that if $A=B$, then $f(A) = f(B)$ for all functions $f$, this proves what you want.

 

[Thinker301]

That was the coolest thing I have heard all day, and I am watching Doctor Who right now.

EDIT : Wait doesn't that create a contradiction. If you do what you did at the top. The axiom would say that they are equal, but the way we defined the function(*) above would contradict that.

 

[adjacent]

Awesome micromass, thanks

 

[verty]

Sure, it should be. But that's why you need axioms.

I like to think in terms of substitution rules, given formula A, substitute formula B. For example, "a + S(b) <=> S(a) + b", "a + 0 <=> a".

Thinking like this, what would one substitute in place of "a = b", probably something almost identical: "a = b <=> a <equals> b". Equality is very fundamental, whatever one substitutes will look essentially similar. So for me it's not an axiom, it's just a fundamental concept that is part of the language. We know what equality is and = is the symbol we use for that concept.

But something like "a + (-a) = 0" could be an axiom which I suppose could be written in this odd substitution style: "a + (-a) <=> 0".

So my response to this thread is that the proof of "if a = b then a+c = b+c" is, by inspection, QED.

 

[verty]

Here is a simple example, consider the operation $*$ on $\mathbb{Q}$ defined as follows:

$$\frac{a}{b}*\frac{c}{d} = \frac{a+c}{b+d}$$

This satisfies all of Spivak's axioms for $+$ (associative, zero element, inverses, commutative).

But then for example, we have $A = \frac{0}{1}$, $B=\frac{0}{2}$ and $C=\frac{1}{1}$. Certainly, $A=B$, but

$$A*C = \frac{0}{1}*\frac{1}{1} = \frac{0+1}{1+1} = \frac{1}{2}$$

and

$$B*C = \frac{0}{2}*\frac{1}{1} = \frac{0+1}{2+1} = \frac{1}{3}$$

So $A=B$, but $A*C\neq B*C$. So there are funny examples of "operations" that do not satisfy this. What goes wrong of course is that the function $f(A) = A*C$ is not a function, since $A=B$ does not imply here that $f(A) = f(B)$.

So Spivak should have added an extra axiom that $+$ is a function. Together with the axiom of equality that if $A=B$, then $f(A) = f(B)$ for all functions $f$, this proves what you want.

Yes, I see there is a clash of views or approaches here. I'll need to think about this.

 

[micromass]

I like to think in terms of substitution rules, given formula A, substitute formula B. For example, "a + S(b) <=> S(a) + b", "a + 0 <=> a".

Thinking like this, what would one substitute in place of "a = b", probably something almost identical: "a = b <=> a <equals> b". Equality is very fundamental, whatever one substitutes will look essentially similar. So for me it's not an axiom, it's just a fundamental concept that is part of the language. We know what equality is and = is the symbol we use for that concept.

But something like "a + (-a) = 0" could be an axiom which I suppose could be written in this odd substitution style: "a + (-a) <=> 0".

So my response to this thread is that the proof of "if a = b then a+c = b+c" is, by inspection, QED.

http://en.wikipedia.org/wiki/First_order_logic#Equality_and_its_axioms

 

[verty]

http://en.wikipedia.org/wiki/First_order_logic#Equality_and_its_axioms

There are several different conventions for using equality (or identity) in first-order logic. The most common convention, known as first-order logic with equality, includes the equality symbol as a primitive logical symbol which is always interpreted as the real equality relation between members of the domain of discourse, such that the "two" given members are the same member. This approach also adds certain axioms about equality to the deductive system employed. These equality axioms are...

It says that this first convention interprets equality as real equality and adds certain axioms about equality to the deductive system employed. It goes on to say that the other convention leaves equality to be defined by substitution rules, that is, definitions. So we either accept equality as given (axiomatized) real equality or define it for our objects.

I'll accept that real equality is axiomatized. This seems to be the only difference. My proof changes slightly from "by inspection, QED" to "by an axiom of real equality, QED". But philosophically I don't like this.

I'd better explain this. I don't like it because the deductive system I used is the one in my head and I used inspection to prove it to myself. That is, I proved it by inspection, not by an axiom.

This (point of view) is what Quine calls psychologism, one of the two dogmas of empiricism. Or not, actually, I can't remember now, too many years ago. Anyway, I'm tired of this.

 

[verty]

The bottom line is, I concede. I liked the old style of math where there could be sets of elephants or whatever, math was about real things, equality was real equality, numbers were equivalence classes, etc. Now it's all pushing symbols, no insight anymore. Math has lost something, even though the language is sure to be simpler.

So let it be "by an axiom of equality" or "by the axiom scheme of substituting equals for equals". The ironic thing is, there were controversies in the past about math being about things that didn't exist, be they infinitesimals or infinite numbers or whatever. Rigor and precision aimed to avoid talking about things such as these. Then set theory happened and you could talk about them because they were just sets, the language became freer. I guess the old sense is gone for good. It's now equality as defined or as axiomatized.

What was inane before, this question of whether, if a=b, a+c = b+c, is now posed as a valid question needing to be answered; by becoming less philosophical we've become more pedantic. I would like to say we are learning about numbers, not how numbers are defined. This just seems yucky.

 

[johnqwertyful]

The bottom line is, I concede. I liked the old style of math where there could be sets of elephants or whatever, math was about real things, equality was real equality, numbers were equivalence classes, etc. Now it's all pushing symbols, no insight anymore. Math has lost something, even though the language is sure to be simpler.

So let it be "by an axiom of equality" or "by the axiom scheme of substituting equals for equals". The ironic thing is, there were controversies in the past about math being about things that didn't exist, be they infinitesimals or infinite numbers or whatever. Rigor and precision aimed to avoid talking about things such as these. Then set theory happened and you could talk about them because they were just sets, the language became freer. I guess the old sense is gone for good. It's now equality as defined or as axiomatized.

What was inane before, this question of whether, if a=b, a+c = b+c, is now posed as a valid question needing to be answered; by becoming less philosophical we've become more pedantic. I would like to say we are learning about numbers, not how numbers are defined. This just seems yucky.

This whole post reeks of smugness.

What "old style of math"? Axioms have been around since Euclid! This is the only way math is ever done. If you don't like that, fine. Don't study mathematics. I have no idea how we could possibly study math if not axiomatically.

I have no idea why you think it's pushing symbols and no insight. Look at how much of Spivak is prose vs how much is symbols. There's more words than symbols! I have no idea where you get the idea. I don't even see how that comes from the thread. We were talking about axioms, which try to make precise loose notions we have and formally justify manipulations. This approach does the opposite of what you said.

What has "math lost"? If you're having trouble understanding axiomatic systems, I seriously doubt you have any real experience with mathematics to even make that call. How many papers have you published on mathematics? Do you have a PhD or even a bachelors in math? If not, what makes you think you're qualified to comment on how math changes with time? I'm sure it has, but very few people are qualified to comment on that.

What are "real things"?

 

[verty]

What "old style of math"? Axioms have been around since Euclid! This is the only way math is ever done.

I didn't realize this but Euclid does include axioms such as "the whole is greater than the part" and "things which coincide are equal", that is, axioms of equality. He didn't take it for granted. So I can't speak against that, it is a historical fact.

If you don't like that, fine. Don't study mathematics. I have no idea how we could possibly study math if not axiomatically.

I think we could study math without axioms but that is beside the point. Math isn't studied that way and that's that.

I have no idea why you think it's pushing symbols and no insight.

Why couldn't we just say, let = be equality in the usual sense? Because it's a historical fact that we don't. I get that now.

We were talking about axioms, which try to make precise loose notions we have and formally justify manipulations. This approach does the opposite of what you said.

What I said was, if it's all axioms and no insight, it's pushing symbols. I think your point is that it's not "all" that, Spivak has more words than prose, clearly there is insight there. Yes, clearly there is insight to be gained by pushing symbols. But proving what this thread asks to prove has I think nothing to do with insight. So let's stay focused on that.

If you're having trouble understanding axiomatic systems, I seriously doubt you have any real experience with mathematics to even make that call.

I learned axiomatic set theory and a ton of math without every understanding it.

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There was more to this post but this is not the place. I want this not to continue.

 

[johnqwertyful]

I think we could study math without axioms but that is beside the point. Math isn't studied that way and that's that.

Try to understand why it's necessary. It's not because "it's how things are done". It's because without axioms, there's no way to do math. Math IS axioms.

Why couldn't we just say, let = be equality in the usual sense?

What does "in the usual sense" even mean?

What I said was, if it's all axioms and no insight, it's pushing symbols.

Why are axioms and insight mutually exclusive?

 

[verty]

Final thought: mathematics is the science of quantity. Whatever flows from that is acceptable (if it's good for science).

 

[johnqwertyful]

Even that which is not good for science is acceptable. ;)

 

[micromass]

OK, I think this thread is done now.