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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

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  • $\begingroup$ 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? $\endgroup$ Dec 15 '13 at 21:55
  • $\begingroup$ No, I just put it in this order. He shows them P1-P2-P3. Ill change it :) $\endgroup$
    – user108343
    Dec 15 '13 at 21:57
  • $\begingroup$ You wrote "Now, he tries to prove P2" so the book is trying to prove P2 right? $\endgroup$ Dec 15 '13 at 22:02
  • $\begingroup$ Yeah that's what he tries to do. $\endgroup$
    – user108343
    Dec 15 '13 at 22:03
  • $\begingroup$ and he uses P1 and P3 to prove it, so P1 and P3 together imply P2 so why you said no? $\endgroup$ Dec 15 '13 at 22:05
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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.

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  • $\begingroup$ 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! $\endgroup$
    – user108343
    Dec 15 '13 at 22:31
  • $\begingroup$ Oops! typo! I meant $x$ ! $\endgroup$ Dec 15 '13 at 22:32
  • $\begingroup$ OK, And for my second comment ? Can you explain me ? Thanks $\endgroup$
    – user108343
    Dec 15 '13 at 22:34
  • $\begingroup$ We are not trying to prove P1, P2 nor P3. We are trying to prove that there is a unique additive identity in $\mathbb{R}$. $\endgroup$ Dec 15 '13 at 22:37
  • $\begingroup$ 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. $\endgroup$ Dec 15 '13 at 22:41

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