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Old 2021-04-18, 13:08   #1
drkirkby
 
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Default How do I prove (a+b)^2=a^2+2ab+b^2 from axioms?

I studied engineering at university, and whilst I have done quite a bit of mathematics, I have never formally studied number theory. The following is not homework, but study for self interest.

I am trying to get a grip on number theory, and have the book “Elementary Number Theory and its applications” by Rosen (2nd edition). The book starts with the basic properties of the integers, which it calls axioms. Now all these are blindingly obvious to me, as I have always accepted that a+b=b+a and other similar simple things without question. However, I do appreciate the advantage of having a set of axioms where these things are formally stated.

I have copied a page from the book, which has the axioms. The book has some questions, one of which is to prove (a+b)^2=a^2+2ab+b^2, using the axioms for the integers. Now of course I know (a+b)^2 is the same as (a+b)(a+b), but there’s nothing about powers in the axioms, so I don’t know how to prove the question, without resorting to something that is blindly obvious, but does not follow from any of the axioms as far as I can see.

Dave
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Old 2021-04-18, 13:15   #2
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You'll have to use the distributive laws (and rearrange terms using the axiom you mentioned).

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Old 2021-04-18, 14:02   #3
drkirkby
 
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I am mot following you. The distributive law states (a+b)c = ac + bc. But I don’t see how to get (a+b)^2 into a format that I can apply the distributive law. I don’t even see a way to say what a^2 is, without basing it on knowledge not given by the axioms. Obviously I know a^2 is a times a, but that doesn’t follow from the axioms.
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Old 2021-04-18, 14:14   #4
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Quote:
Originally Posted by drkirkby View Post
I am mot following you. The distributive law states (a+b)c = ac + bc. But I don’t see how to get (a+b)^2 into a format that I can apply the distributive law. I don’t even see a way to say what a^2 is, without basing it on knowledge not given by the axioms. Obviously I know a^2 is a times a, but that doesn’t follow from the axioms.
(a+b)^2 means A*B where A=B=a+b
(a+b)*(a+b)=a*(a+b)+b*(a+b) with right distribution
a*(a+b)+b*(a+b)=a^2+a*b+b*a+b^2 with two applications of left distribution
The desired result follows immediately, since for integers commutativity of multiplication holds: a*b=b*a

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Old 2021-04-18, 14:31   #5
Nick
 
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Quote:
Originally Posted by drkirkby View Post
Obviously I know a^2 is a times a, but that doesn’t follow from the axioms.
What's missing is definition of the notation. We can define it inductively as follows:
Take any integer a.
We define \(a^0=1\) and, for each non-negative integer n, define \(a^{n+1}=a^n\cdot a\).
The usual laws of indices can then be proved by induction (using the axioms).
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Old 2021-04-18, 14:43   #6
LaurV
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Yep, as they said.

You should not be confused by powering, that is just defined as a repeated multiplication, same way as the multiplication is defined as a repeated addition. That is, the "a*b" is just a shorthand writing for "a+a+a+...+a", where "a" appears "b" times. In the same way, "a^b" is just a shorthand writing for "a*a*a*...*a", where "a" appears "b" times. In fact, you can extend this in both directions, like for example, everything starts with a "unit" (a matchstick) which you can stack up to make quantities (called "numbers"). The "stacking up" is called incrementing, and the number "a" is just a repeated incrementation i.e. 1+1+1+...+1. Then, "addition", a+b, is just "a, incremented b times", i.e. a+1+1+...+1, where 1 appears "b" times. (this is in fact useful when you "prove" commutativity of addition and multiplication, which sometimes is not given as "axiom" (usually, operations are not commutative, but the "numbers" are a very special "category" of things, or "group" for which the commutativity is true, see abelian groups). (for extension on the other direction, see the arrow notation).

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Old 2021-04-18, 16:14   #7
drkirkby
 
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Thank you everyone. It makes sense now.

I hope you don't mind if I ask other similar questions as I work my way though this (or similar) book(s).

Dave

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Old 2021-04-19, 11:19   #8
xilman
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Quote:
Originally Posted by drkirkby View Post
Thank you everyone. It makes sense now.

I hope you don't mind if I ask other similar questions as I work my way though this (or similar) book(s).

Dave
Now try proving it as in Russell & Whitehead's Principia Mathematica.

It took them 379+86 = 465 pages to get to the point where they proved 1+1=2. The proof was almost complete by p379 but they had not yet defined addition at that point.

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Old 2021-04-19, 11:36   #9
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https://en.wikipedia.org/wiki/Principia_Mathematica

A very soporific read!
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Old 2021-04-20, 07:27   #10
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If you want to see a computer-assisted proof, Metamath's version is at http://us.metamath.org/mpeuni/binom2.html. The associated program can blow up that heavily condensed proof all the way back to its own axiom set, which would be more like the Principia version than what you asked for.
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Old 2021-04-22, 15:41   #11
dlsilver06
 
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Use the distributive law (a+b).c = ac + bc
Replace c by (a+b): (a+b).(a+b) = a (a+b) + b (a+b)
Then = a2 + ab + ba +b2
= a2 + 2ab + b2
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