Simple Proof question

Question

Image : http://postimg.org/image/dkn0d5uen/ I'm studying Spivak's calculus and I have a really simple question :

I'm only in the first chapter on "The basic properties of numbers"

So far, we have the following propostion

P1 : (a+b)+c=a+(b+c)

P2 : a+0=0+a=a

P3 : a+(-a)=(-a)+a=0

Now, he tries to prove P2 (He doesn't do it for P3, so it's granted) He also says :

"The proof of this assertion involves nothing more than subtracting a from both sides of the equation, in other word, adding -a to both sides." Now, that I understand

"as the following detailled proof shows, all three properties P1-P3 must be used to justify this operation." That I don't understand. First, how can you use a proof of something you haven't proven ? Second, when he says all three properties to justify this operation, he means to substract "a" from both sides, right ? If so, I don't understand how they (properties) can be used ...

He starts with this :

If a+x=a

then (-a)+(a+x)=(-a)+a=0

hence ((-a)+a)+x=0

hence 0+x=0

hence x=0

My comments : For the first line, he starts with the assertion that an equation a+x=a exists. Now, he substract "a" from borth sides and with property 3 the right hand sides equals 0. With property 1 we regroup and cancel with property 3.Now we have 0+x=0 and we subtract zero from both sides to have x=0. Where is property 2 used ? How is subtracting "a" from both sides proven with all three properties ?

Thank you

The question is not clear. Are you trying to say that the book says P1 and P3 implies P2? And if yes, why is this the case? — Sergio Parreiras, Dec 15 '13 at 21:55
No, I just put it in this order. He shows them P1-P2-P3. Ill change it :) — , Dec 15 '13 at 21:57
You wrote "Now, he tries to prove P2" so the book is trying to prove P2 right? — Sergio Parreiras, Dec 15 '13 at 22:02
and he uses P1 and P3 to prove it, so P1 and P3 together imply P2 so why you said no? — Sergio Parreiras, Dec 15 '13 at 22:05
Well, he says that he uses the three of them, that's the problem (Like it says in the quote.) But when he presents the properties of numbers he does it in order , so : P1-P2-P3 So I wouldn't say he imples P2 by using P1-P3 Ill show you a picture of the page if you wan to (If you dont think it's clear.) — , Dec 15 '13 at 22:12
Just a thought : the operation of "cancelling" $a$ in the equality $b+a = c+a$ to deduce $b = c$ requires all three properties P1-P3. Could the author be talking about this operation ? — Guest, Dec 15 '13 at 22:12
Yeah, that's what I thought also. But why does he say in the quote that he used the 3 of them ?? I'm someone who takes every word very seriously.... Unless it's a mistake from him ? (or from me) — , Dec 15 '13 at 22:19

Sergio Parreiras · Accepted Answer · 2013-12-15T22:39:49.900

1

Spivak wants to show that zero is the unique additive identity on $\mathbb{R}$. That is, he want to prove that if we have $a+x=a$ then $x$ must identical to zero. He assumes P1, P2 and P3 to prove this. In particular, he uses P2 in the last step. If $0+x=0$ then using P2 we can conclude that $x=0$ without P2 we can not conclude this.

edited Dec 15 '13 at 22:39

answered Dec 15 '13 at 22:25

Sergio Parreiras

3,813

Wait... What do you mean by "then a must identical to zero." He says "a" can represent any number... Also, I've asked this question because I didn't understand how we can use a property that we are trying to prove. Isn't that illogical ? (well, for me it is ...) Thank you! – Dec 15 '13 at 22:31
Oops! typo! I meant $x$ ! – Sergio Parreiras Dec 15 '13 at 22:32
OK, And for my second comment ? Can you explain me ? Thanks – Dec 15 '13 at 22:34
We are not trying to prove P1, P2 nor P3. We are trying to prove that there is a unique additive identity in $\mathbb{R}$. – Sergio Parreiras Dec 15 '13 at 22:37
We will not try to prove P1, P2, and P3 because we can imagine other "systems of numbers" where one or more of these properties are not true. P1, P2 and P3 are axioms we take them as given. – Sergio Parreiras Dec 15 '13 at 22:41
Just to clarify more: Spivak wants to show that we can not have two different elements, call them $0$ and $\hat 0$ such that both of them are additive identities $x+0=x$ and $x+\hat 0=x$ for all $x$. But we can imagine other sets (where one of the P1, P2 and P3 do not hold) where this is possible. – Sergio Parreiras Dec 15 '13 at 22:43
Yeah, I think that's it. I don't why I didn't understand it that way... Lastly, when he says : "as the following detailled proof shows, all three properties P1-P3 must be used to justify this operation." This operation means what exactly ? – Dec 15 '13 at 22:51
"this operation" is the proof of $\left[ a+x=a \Rightarrow x=0 \right]$ – Sergio Parreiras Dec 15 '13 at 23:00
Ok, thank you for your help ! I really messed up in understanding the text ^^ – Dec 15 '13 at 23:05

Simple Proof question

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