# Spivak's Calculus, p3: a = b implies a + c = b + c

1. Jul 24, 2010

### chollapete

1. The problem statement, all variables and given/known data

Not a specific exercise per se; just a question from reading the text. In working the proof shown at the bottom of page 4 (2nd edition), I realized that the step "If a + x = a,
then (-a) + (a + x) = (-a) + a" could not be justified by any of Postulates 1, 2, or 3. At this point, that is all we have to work with.

The missing postulate is: If a = b, then a + c = b + c.

2. Relevant equations

Postulate 1: If a, b, and c are any numbers, then a + (b + c) = (a + b) + c.
Postulate 2: If a is any number, then a + 0 = 0 + a = a.
Postulate 3: For every number a, there is a number -a such that a + (-a) = (-a) + a = 0.

Also, he assumes that given numbers a and b, a+b and ab exist.

3. The attempt at a solution

If a = b,
then a + 0 = b; (P2)
hence a + (c + (-c)) = b; (P3)
hence (a + c) + (-c) = b; (P1)
hence ((a + c) + (-c)) + c = b + c; ??? At this point I believe I need the "missing postulate".
hence (a + c) + ((-c) + c) = b + c; (P1)
hence (a + c) + 0 = b + c; (P3)
hence a + c = b + c. (P2)

Any ideas on how to be rigorous here?

P.S. First post on this forum. :-) Can anyone clue me in on how to use TeX code? Thanks.

2. Jul 25, 2010

### sponsoredwalk

1: a + x = a

2: (a + x) + (-a) = (a + (-a))

3: (a + (-a)) + x = (a + (-a))

4: x = 0

This is exactly what you wrote but you didn't recognise that (a + (-a)) is just another way of writing the number zero, as defined by postulate 3. You see how the brackets link two numbers at a time, i.e. in a binary sort of way? Well them two become 1 & represent 1 number. Also, in your attempted solution you made algebraic errors, of sorts. You did something to one side of the equation but left the other side untouched. I know you're able to do this but I don't think Spivak has allowed anything like that yet :tongue2:

As for latex, you use these brackets [tex] [/ tex] <--- but / tex should be /tex

Also, there is a great firefox plugin I use https://addons.mozilla.org/en-US/firefox/addon/4082/ [Broken] that makes it 10 times easier

Last edited by a moderator: May 4, 2017
3. Jul 25, 2010

### chollapete

Thanks for the reply!

What is bothering me is that Spivak doesn't give us a way to add the same number to both sides of an equality. I came up with the following way to do this, but it requires an additional "pre-postulate", that I call (P -1): If a is a number, then a = a.

Then:
If a, b, and c are numbers and a = b,
then (a + c) exists; (P0)
hence (a + c) = (a + c); (P -1)
hence a + c = a + c; (No postulate, but I'm okay accepting this)
hence a + c = b + c. (Supposition)

So, I'm happy given I get to use the postulate that a number equals itself. :-) Unless I'm making an obvious error, I can see why Spivak didn't think this important enough to make explicit.

I'm a little concerned that the last step makes it a fallacy of begging the question, but I don't think so. Maybe someone could comment about this.

4. Jul 25, 2010

### sponsoredwalk

Spivak does actually give you a way to do that when he is talking about P1.
Just remember that anything you do on one side of an equation must be done
on the other side in order to ensure the equality holds, i.e.
make sure "=" is a true statement.

In the 3rd edition, my one, Spivak says;

"For example, we know from P1 that

(a + b) + c = a + (b + c)

...

so you can see that if a = b then your proof will work fine.

I've just added that little bit to show you can go further, but it's the right idea!

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