Spivak: Basic properties of numbers

In summary, the conversation discusses the necessity of a step in Spivak's Calculus proof and the importance of rigor in mathematics. The step in question is seen as a justification of x + x = 2x and is necessary in the context of the properties previously introduced. The conversation also touches on the importance of proving everything from axioms and the role of proof in mathematics. Finally, the conversation concludes with a discussion of the uniqueness of the additive inverse. The conversation is driven by the question of why the step in bold was necessary in the proof and the benefits of rigor in mathematics.
  • #1
Saladsamurai
3,020
7
Hello all! :smile:

I am (painfully) going through the first chapter of Spivak's Calculus. At one point he introduces the property: if a, b, and c are any numbers, then [itex]a\cdot(b+c) = a\cdot b +a\cdot c[/itex]. He then uses this property in an example in which he shows that the only time that [itex] a - b = b- a[/itex] is when a = b.[itex]a - b = b - a [/itex]
[itex] \Rightarrow (a - b) + b = (b - a) + b = b + (b - a) [/itex]
[itex]\Rightarrow a = b + b - a [/itex]
[itex]\Rightarrow a + a = (b + b - a) + a = b + b [/itex]
[itex]\mathbf{ \Rightarrow a \cdot (1 + 1) = b \cdot (1 + 1) } [/itex]
[itex]\Rightarrow a = b [/itex]I am unsure of why the step in bold was necessary? Can someone elaborate as to why he included this step? In fact, he said it was necessary to include this step.
 
Last edited:
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  • #2
Well, how would you do it without the bold step??
 
  • #3
That step is basically justification of x + x = 2x.
 
  • #4
It would help greatly if you explained why you don't think it's necessary. How would you complete the proof?
 
  • #5
Kevin_Axion said:
What step in bold?
See edit.

micromass said:
Well, how would you do it without the bold step??

Hmmm. Good question :smile: It just seems weird to do that step (to me). I guess I see where this is going:

I would have said that a + a = b + b simply means 2a = 2b but I suppose that is what he is driving at? That in terms of the properties he has just introduced a + a = a (1 + 1) which is 2a.EDIT: Wow: You guys pounced all over this one :smile: Better watch myself roaming into this forum.
 
  • #6
Saladsamurai said:
See edit.



Hmmm. Good question :smile: It just seems weird to do that step (to me). I guess I see where this is going:

I would have said that a + a = b + b simply means 2a = 2b but I suppose that is what he is driving at? That in terms of the properties he has just introduced a + a = a (1 + 1) which is 2a.


EDIT: Wow: You guys pounced all over this one :smile: Better watch myself roaming into this forum.


He's trying to be rigorous and follow from the propositions he proved before.
 
  • #7
Saladsamurai said:
See edit.



Hmmm. Good question :smile: It just seems weird to do that step (to me). I guess I see where this is going:

I would have said that a + a = b + b simply means 2a = 2b but I suppose that is what he is driving at? That in terms of the properties he has just introduced a + a = a (1 + 1) which is 2a.


EDIT: Wow: You guys pounced all over this one :smile: Better watch myself roaming into this forum.

Indeed, he just wants to do 2a=2b and eliminate the 2. But he didn't introduce 2 yet, so he calls it 1+1.

The argument should be more rigorous though. He should also show that [itex]1+1\neq 0[/itex]. You should really prove everything from the axioms. Don't use anything else.
 
  • #8
micromass said:
Indeed, he just wants to do 2a=2b and eliminate the 2. But he didn't introduce 2 yet, so he calls it 1+1.

The argument should be more rigorous though. He should also show that [itex]1+1\neq 0[/itex]. You should really prove everything from the axioms. Don't use anything else.

I think that this is in an exercise somewhere. But seriously, why should this be proved? Of what benefit does this offer? And moreover, once one shows that [itex]1 + 1 \ne 0[/itex] then do I need to show that [itex] 1 + 1 + 1 \ne 0[/itex] and that [itex]1 +1 +1 +1 \ne 0 [/itex]? Where does it end?

Also, I am being serious, not facetious. I have an engineer's background in math, so for me, there is really no need for all that rigor.
 
  • #9
Saladsamurai said:
I think that this is in an exercise somewhere. But seriously, why should this be proved? Of what benefit does this offer? And moreover, once one shows that [itex]1 + 1 \ne 0[/itex] then do I need to show that [itex] 1 + 1 + 1 \ne 0[/itex] and that [itex]1 +1 +1 +1 \ne 0 [/itex]? Where does it end?
It ends whenever you stop...

Proof is the backbone of mathematics. It let's us know that we're correct and gives us a glimpse into why. Whether you need to prove that [itex]1 + 1 + 1 \ne 0[/itex] or not depends on how you go about your proofs. For instance, you can use mathematical induction to prove that no number of additions of the multiplicative identity of the real numbers will equal the additive identity. It's important to note that there are other number systems where this is not true: where you can add 1 a number of times to get 0, so it's nice to prove that this can't be the case for real numbers...

Also, I am being serious, not facetious. I have an engineer's background in math, so for me, there is really no need for all that rigor.
I'm sure we all get that. I know I do. Do you understand that there is a place for this kind of rigor?

Incidentally, why do you find going through Spivak "painful?"
 
  • #10
Just out of curiosity, how does one prove [itex]1 + 1 \neq 0[/itex]?
 
  • #11
Yuqing said:
Just out of curiosity, how does one prove [itex]1 + 1 \neq 0[/itex]?

You define what addition is...
 
  • #12
Is there an axiom that requires the additive inverse to be unique?
 
  • #13
Kevin_Axion said:
You define what addition is...

So we basically define 1 + 1 = 2 and that 2 does not equal 0?
 
  • #14
I like Serena said:
Is there an axiom that requires the additive inverse to be unique?
No but you can prove this. For instance:

Suppose we have a real number a and two additive inverses, b and c. By definition, a + b = 0 and a + c = 0. It follows that a + b = a + c. Therefore b = c.
QED.
 
  • #15
Yuqing said:
Just out of curiosity, how does one prove [itex]1 + 1 \neq 0[/itex]?
Actually, I don't think you need to prove each specific example. By definition, [itex]1 + 1 = 2 \neq 0[/itex]. However, while you can do this for any number of additions of the multiplicative identity, this doesn't prove that it's true for any number of additions. This would require some kind of proof...

Personally, with this, I would use induction and the ordering of the reals...

I suspect that the reason Spivak didn't simply use a + a = 2a is because you'd derive that from the distributive property like so: a + a = (1 + 1) a = 2a. So for him, he's actually skipping a step by simply cancelling out (1 + 1)...
 
  • #16
Saladsamurai said:
I think that this is in an exercise somewhere. But seriously, why should this be proved? Of what benefit does this offer? And moreover, once one shows that [itex]1 + 1 \ne 0[/itex] then do I need to show that [itex] 1 + 1 + 1 \ne 0[/itex] and that [itex]1 +1 +1 +1 \ne 0 [/itex]? Where does it end?

Also, I am being serious, not facetious. I have an engineer's background in math, so for me, there is really no need for all that rigor.

Jocko Homo said:
No but you can prove this. For instance:

Suppose we have a real number a and two additive inverses, b and c. By definition, a + b = 0 and a + c = 0. It follows that a + b = a + c. Therefore b = c.
QED.

Spivak is the route for many into formal maths.

Saladsamurai, as an engineer you will have doubtless used the technique of practicing on a simple known and often trivial example before tackling more complicated stuuf or simply as a verification of a technique.

I suggest taking Spivak's and JH examples in this vein.

go well
 
  • #17
Jocko Homo said:
Actually, I don't think you need to prove each specific example. By definition, [itex]1 + 1 = 2 \neq 0[/itex]. However, while you can do this for any number of additions of the multiplicative identity, this doesn't prove that it's true for any number of additions. This would require some kind of proof...

So is it actually defined that [itex]1 + 1 = 2 \neq 0[/itex] for most constructions of the reals? I only ask because in most constructions of the reals that I've seen this point seems largely neglected. In fact, I haven't bothered to think about it before now.
 
  • #18
Studiot said:
Spivak is the route for many into formal maths.

Saladsamurai, as an engineer you will have doubtless used the technique of practicing on a simple known and often trivial example before tackling more complicated stuuf or simply as a verification of a technique.

I suggest taking Spivak's and JH examples in this vein.

go well

Hi studiot :smile: Yes you are correct. I am taking a bit of a detour now that I am done with my formal engineering studies. It's just that Spivak has a lot of little side notes like the one that spawned this thread. I can certainly move forward without proving that [itex]1 + 1 \ne 0[/itex]. But as I move forward, I wish to know what motivates him to suggest that this is even worth worrying about (if only once).

I am still not sure if I have gotten an answer to that question. Both Spivak and micromass found it non rigorous that the proof in post #1 did not include a step in which it was shown that [itex]1 + 1 \ne 0[/itex]. I suppose I am beginning to see it though.

If 1+1=0 was true, then the step in bold could not be multiplied on both sides by (1+1)-1 and hence the proof would not hold.

Just getting used to looking at things from a different perspective. :wink:
 
  • #19
It would be undefined.
 
  • #20
Saladsamurai said:
Hi studiot :smile: Yes you are correct. I am taking a bit of a detour now that I am done with my formal engineering studies. It's just that Spivak has a lot of little side notes like the one that spawned this thread. I can certainly move forward without proving that [itex]1 + 1 \ne 0[/itex]. But as I move forward, I wish to know what motivates him to suggest that this is even worth worrying about (if only once).

I am still not sure if I have gotten an answer to that question. Both Spivak and micromass found it non rigorous that the proof in post #1 did not include a step in which it was shown that [itex]1 + 1 \ne 0[/itex]. I suppose I am beginning to see it though.

If 1+1=0 was true, then the step in bold could not be multiplied on both sides by (1+1)-1 and hence the proof would not hold.

Just getting used to looking at things from a different perspective. :wink:

The thing is that mathematicians like Spivak try to prove statements without assuming anything but the axioms. Spivak has presented a number of axioms (about 15 or so), and we should prove everything from these 15 axioms. We cannot use anything else. That is what Spivak wants to do.

Of course, you won't need to prove that [itex]1+1\neq 0[/itex] in order to proceed. You will still understand the theory. In fact, maybe you'll even find the book easier if you don't prove this. However, proving such a thing is benificial because it gives you acquaintance with easy proofs.

I guess that it's a point-of-view thing. Spivak assumes nothing but the axioms. He assumes nothing else. That's his point-of-view. I hope that explains why Spivak does the things that he does...
 
  • #21
micromass said:
The thing is that mathematicians like Spivak try to prove statements without assuming anything but the axioms. Spivak has presented a number of axioms (about 15 or so), and we should prove everything from these 15 axioms. We cannot use anything else. That is what Spivak wants to do.

Of course, you won't need to prove that [itex]1+1\neq 0[/itex] in order to proceed. You will still understand the theory. In fact, maybe you'll even find the book easier if you don't prove this. However, proving such a thing is benificial because it gives you acquaintance with easy proofs.

I guess that it's a point-of-view thing. Spivak assumes nothing but the axioms. He assumes nothing else. That's his point-of-view. I hope that explains why Spivak does the things that he does...

Hi micromass :smile: Yes. This is what I am gathering from what I have read, and believe it or not I like it! And you are exactly correct when you say "However, proving such a thing is benificial because it gives you acquaintance with easy proofs."

This is exactly why I am bothering with these things. I am trying out my hand at mathematics from a math point of view. As an engineer, I have always asked a lot of questions that most engineers (my peers) consider trivial. I like to ask "how can you be sure? Where is the proof?" I guess that is what Spivak does too and why I am taking a liking to him.

But to answer another poster's question: It is a little 'painful' because I am not used to it. But that does not mean I do not enjoy it :wink:

I hope that the folks here in the math forums won't mind me starting more threads like this. I like to ask lots of questions, especially those regarding the motivation of a particular subject.

Thanks for all of your time :smile: I think my question was answered.
 
  • #22
Saladsamurai said:
I hope that the folks here in the math forums won't mind me starting more threads like this.

Not at all! Ask at will :biggrin:
 
  • #23
Saladsamurai said:
But to answer another poster's question: It is a little 'painful' because I am not used to it. But that does not mean I do not enjoy it :wink:
Double negative... tricky...

I'm glad to hear you're... getting some enjoyment from it!

I hope that the folks here in the math forums won't mind me starting more threads like this. I like to ask lots of questions, especially those regarding the motivation of a particular subject.
I'd be shocked if anyone here would "mind," especially given your growing enthusiasm for the subject! You'll encounter some amazing and unbelievable results in Spivak that you'd never encounter as an engineer...
 

1. What is the purpose of Spivak's "Basic Properties of Numbers"?

Spivak's "Basic Properties of Numbers" is a mathematical textbook designed to introduce readers to the fundamental concepts and properties of numbers, including integers, rational numbers, and real numbers.

2. Who is the target audience for this book?

This book is primarily aimed at high school or college students who are new to the study of mathematics and want to develop a strong understanding of the basic properties of numbers.

3. What topics are covered in Spivak's "Basic Properties of Numbers"?

The book covers a wide range of topics, including the properties of integers, rational numbers, real numbers, and complex numbers. It also delves into topics such as prime numbers, divisibility, and modular arithmetic.

4. Is this book suitable for self-study?

Yes, this book is suitable for self-study as it includes clear explanations, examples, and exercises to help readers develop their understanding of the material at their own pace.

5. Are there any prerequisites for reading this book?

Some basic knowledge of algebra and geometry is recommended for readers to fully understand the concepts covered in this book. However, the book does include a brief review of these topics in the beginning chapters.

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