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 2021-04-18, 13:08 #1 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 3·59 Posts 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 Attached Thumbnails   Last fiddled with by drkirkby on 2021-04-18 at 13:15 Reason: Add photograph
 2021-04-18, 13:15 #2 paulunderwood     Sep 2002 Database er0rr 1110010100002 Posts You'll have to use the distributive laws (and rearrange terms using the axiom you mentioned). Last fiddled with by paulunderwood on 2021-04-18 at 13:15
 2021-04-18, 14:02 #3 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 3·59 Posts 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.
2021-04-18, 14:14   #4
paulunderwood

Sep 2002
Database er0rr

1110010100002 Posts

Quote:
 Originally Posted by drkirkby 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

Last fiddled with by paulunderwood on 2021-04-18 at 14:21

2021-04-18, 14:31   #5
Nick

Dec 2012
The Netherlands

7·239 Posts

Quote:
 Originally Posted by drkirkby 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).

 2021-04-18, 14:43 #6 LaurV Romulan Interpreter     Jun 2011 Thailand 25·5·59 Posts 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). Last fiddled with by LaurV on 2021-04-18 at 14:46
 2021-04-18, 16:14 #7 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 3·59 Posts 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 Last fiddled with by drkirkby on 2021-04-18 at 16:15
2021-04-19, 11:19   #8
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

11·971 Posts

Quote:
 Originally Posted by drkirkby 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.

Last fiddled with by xilman on 2021-04-19 at 11:20 Reason: Minor tweak

 2021-04-19, 11:36 #9 paulunderwood     Sep 2002 Database er0rr 24×229 Posts https://en.wikipedia.org/wiki/Principia_Mathematica A very soporific read!
 2021-04-20, 07:27 #10 Happy5214     "Alexander" Nov 2008 The Alamo City 22·149 Posts 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.
 2021-04-22, 15:41 #11 dlsilver06   Apr 2020 7 Posts 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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