# Getting into the depths of a proof

1. Nov 21, 2009

### evagelos

For the following proof:

$$(a+b)^2 = (a+b)(a+b) = (a+b)a +(a+b)b =a^2 +b^2 +2ab$$.

Write:

a) a list of theorems and definition involved in the proof.

b) a list of the laws of logic involved in the proof by transforming the theorems and the definitions of the proof into the statements of the proof

2. Nov 22, 2009

### tiny-tim

Hi evagelos!

You first!

3. Nov 22, 2009

### tauon

well, the main thing you use here is distributivity of multiplication over addition and the fact that "=" is a symmetrical relation, and rules of general calculation.

for example

the first thing you do is:

$${(a+b)}^2={(a+b)(a+b)}$$ because that's how multiplication of a number with itself a a finite amount of times is defined: $${x*x*x*...*x \ n \ times \ is} \ x^n$$ and because "=" is a symmetrical relation (meaning that if x=y then y=x)

than you use the fact that multiplication is distributive over addition (and again that "=" is symmetrical) to expand the expression and get (a+b)(a+b)=(a+b)a+(a+b)b and so on and so forth... that's for a) and for b) you list the rules of deduction that "allow" those computation to be made (such as if $${(P\rightarrow S)and(P)$$ is true, then S is true).

hope you get the idea. :)

Last edited: Nov 22, 2009
4. Nov 22, 2009

### evagelos

Should we say that the definition involved is:

for all ,nεN , xεR : $$x^n = xx^{n-1}$$ ?

5. Nov 22, 2009

### tauon

you must say that for any set $$S$$ closed under an associative algebraic operation "$$\ast$$" i.e. ($$S , \ast$$) is a semigroup,

$$\forall x \in S$$ , $$n \in \mathbb{N}$$ , $$x^n = \underbrace{x \ast x \ast ... \ast x}_{n \ times}$$ .

that the algebraic operation is associative is very important (the definition doesn't hold if the operation is not associative)...
but x needn't necessarily be a real number. for example the definition "works" for ($$\mathbb{Z}_n , \cdot$$) as well.

Last edited: Nov 22, 2009
6. Nov 22, 2009

### tiny-tim

erm

actually all i was expecting you to say was that it's a definition … x2 = xx

ok … now (a+b)(a+b) = (a+b)a + (a+b)b ?​

7. Nov 22, 2009

### evagelos

should we say now for that statement responsible is the axiom : for all ,x,y,z : z(x+y) = zx + zy??

8. Nov 22, 2009

### tiny-tim

Yes (well, the other way round ).

(and that's called the "…" law ? )

ok, now (a+b)a + (a+b)b = a2 + b2 + 2ab ?

9. Nov 22, 2009

### evagelos

No .the other way round ,because if you put ; z= a+b , x=a ,y = b we have:

z(x+y) = (a+b)a + (a+b)b and to use the same distributive law we must use commutativity and change ,(a+b)a + (a+b)b to a(a+b) + b(a+b) ,and using again distributivity we have:

$$a^2 + ab + ab + b^2$$

Now we must prove that : ab + ab = 2ab

10. Nov 22, 2009

### tiny-tim

That's easy … ab + ab = ab(1) + ab(1)

ok, now part (b)

(if I were you, I'd do that tomorrow :zzz:)

11. Nov 22, 2009

### evagelos

O.K we curry on tomorrow

12. Nov 23, 2009

### evagelos

O.K ,you start first now.

13. Nov 23, 2009

### tiny-tim

hmm … logic isn't my strong point.

ok, well I suppose it must involve the principle that things which are equal to a third thing are equal to each other?

phew!

14. Nov 23, 2009

### evagelos

By applying two times the law of Universal Elimination on the definition:

for all, nεN ,xεR : $$x^n = xx^{n-1}$$ ,where we put n=2 and x = a+b

we have the 1st statement of the proof;

$$(a+b)^2 =(a+b)(a+b)$$

Now what theorem or axiom and what law, the application of which will give us the 2nd statement of the proof i.e

(a+b)(a+b) = (a+b)a + (a+b)b,
do we have to use??

15. Nov 23, 2009

### tauon

I'm not exactly sure how things work around here (if only homework helpers should post or something like that... if it wasn't appropriate for me to post I apologize but...).

Now, I think part b) asks for logical rules of inference (such as modus ponens, reduction ad absurdum, introduction rules etc) used when proving the statement...

I noticed that the depths of this proof "oughtn't" be too rigorous (silly me) but "things which are equal to a third thing are equal to each other" while correct (it follows from the fact that "=" is an equivalence relation -the identity relation-, or since we're talking about logic, from the fact that the equality predicate is transitive), is not a rule of inference.

anyway, I think I intruded enough here, so I'll take my leave.

Last edited: Nov 23, 2009
16. Nov 23, 2009

### evagelos

You are right it is a theorem concerning the predicate of equality