Proof using the Axioms of Addition and Multiplication

[rad0786]

So does everybody know the Axioms of Addition and Multiplication?
They are too long to type, but they are listed:
A1, A2, A3, A4, A5, M1, M2, M3, M4, M5 and the distributive law, DL.

anyways, I want to prove:

1. (-x)y = -(xy) and 2. (-x)(-y) = (xy) using ONLY the axioms of additon and multiplication.

Can somebody please help start me off, I've spent an hour on this and am getting now where :grumpy:

1. seems quite elementry to me.
(-x)y = (-1)(x)y by M4
(-x)y = (-1)(xy) by M3
(-x)y = -(xy) by M3 again

how does that sound?
=

 

[arildno]

Hmm.. I didn't know that (-x)=(-1)x was an axiom.
You should prove that one, and all your troubles vanish.

 

[honestrosewater]

Well, (-x)y + xy = (-x + x)y = ...

 

[quasar987]

I don't know what you call M1,2,3,4, etc but here's how I'd do it:

1° Prove that -x = (-1)(x)

It follows that

2° (-x)y = [(-1)(x)]y = (-1)[(x)(y)] (associativity) = -(xy) (by 1° again)

edit: I like honestrose's way better, but it's good to prove once and for all that -x = (-1)(x).

 

[rad0786]

I don't know what you call M1,2,3,4, etc but here's how I'd do it:

1° Prove that -x = (-1)(x)

It follows that

2° (-x)y = [(-1)(x)]y = (-1)[(x)(y)] (associativity) = -(xy) (by 1° again)

edit: I like honestrose's way better, but it's good to prove once and for all that -x = (-1)(x).

Prove that -x = (-1)(x) : That is a theorm in the textbook, and the proof is their with it.

x + (-1)x = x + x(-1)by M2
=x1 + x(-1) by M4
x[1+(-1)]
x0

about honestrose's way,(-x)y + xy = (-x + x)y = ... (0)y = 0. Is this what he was trying to show as the proof?

Well, (-x)y + xy = (-x + x)y = ...

 

[quasar987]

(-x)y + xy = (-x + x)y (distributivity) = (0)y = 0. Hence (-x)y is the additive inverse of xy, which we note -(xy).

 

[honestrosewater]

about honestrose's way,(-x)y + xy = (-x + x)y = ... (0)y = 0. Is this what he was trying to show as the proof?

Right, and if a + b = 0, then b = ?

I'm a she, by the bye.

 

[rad0786]

Right, and if a + b = 0, then b = ?

I'm a she, by the bye.

if a + b = 0 then b = -a

 

[honestrosewater]

So if (xy) + (-x)y = 0, then (-x)y = ?

What do you have for (2)?

 

[rad0786]

So if (xy) + (-x)y = 0, then (-x)y = ?

What do you have for (2)?

if (xy) + (-x)y = 0, then (-x)y = -(xy) as required.

For (2), i did it a similar way

2. Prove that (-x)(-y) = (xy) using only the axioms...

(-x)(-y) - (xy) = (-x + x)(-y) = 0(-y) = 0
thus, (-x)(-y) - (xy) = 0
therefore, (-x)(-y) = (xy)

how does that sound!? i have a feeling its good!

 

[quasar987]

The general idea is good, and there is nothing you've written that is false.

But it's not detailed enough. You have to justify each and everyone of your = signs by a statement such as "by definition of ..." or "by axiom ..." or "according to theorem ..." or "by the result found in (1)". That's the whole point of proving using the axioms.

 

[rad0786]

The general idea is good, and there is nothing you've written that is false.

But it's not detailed enough. You have to justify each and everyone of your = signs by a statement such as "by definition of ..." or "by axiom ..." or "according to theorem ..." or "by the result found in (1)". That's the whole point of proving using the axioms.

that you quasar987, In my answer, I would write "by M1, by M2, by A4... etc" (refering to the specific axiom.)

I came across another Proving question in the book, (in the same section as the axiom one.)

Prove: If $x \geq 0$ and $x \leq \varepsilon$ for all $\varepsilon > 0$ then $x = 0$

Firstly, I decided to write it up more mathematically.

Prove: If $0 \leq x \leq \varepsilon$ for all $\varepsilon > 0$ then $x = 0$

Now, this statement dosn't even look true to me. Suppose x = 9 and $\varepsilon = 11$, then dosn't the statement hold? But since it says Prove, ill just accept the fact that its true.

So is it possible to prove this using ONLY the axioms? i took the contrapositive and that didnt help :grumpy:

 

[quasar987]

Now, this statement dosn't even look true to me. Suppose x = 9 and $\varepsilon = 11$, then dosn't the statement hold?

The statement is that if $0\leq x \leq \epsilon$ FOR ALL $\epsilon>0$. So if you say x =9, while it is true for e = 11, it is not true for e = 8, hence it is not true for all e>0.

 

[quasar987]

So is it possible to prove this using ONLY the axioms? i took the contrapositive and that didnt help :grumpy:

Try proof by contradiction. Suppose x is not zero. Can you find a number epsilon smaller than x?