# equivalence

• Jun 3rd 2012, 01:24 PM
psolaki
equivalence
are the following equivalent?

0x =0, (-x)y = -xy , (-x)(-y) = xy
• Jun 3rd 2012, 01:34 PM
Plato
Re: equivalence
Quote:

Originally Posted by psolaki
are the following equivalent?
0x =0, (-x)y = -xy , (-x)(-y) = xy

What do you mean by equivalent here?
• Jun 3rd 2012, 07:02 PM
psolaki
Re: equivalence
I mean prove the equivalence between the three statement

I was told to show that : 0x =0 implies (-x)y =-xy and that (-x)y = -xy implies (-x)(-y) = xy and finally that (-x)(-y) = xy implies 0x = 0

And this is an equivalence between the three statement﻿
• Jun 4th 2012, 05:20 AM
emakarov
Re: equivalence
What are other axioms at your disposal? Obviously, if multiplication and negation are arbitrary functions, then 0x = 0 does not imply (-x)y = -xy.
• Jun 4th 2012, 07:06 AM
psolaki
Re: equivalence
Quote:

Originally Posted by emakarov
What are other axioms at your disposal? Obviously, if multiplication and negation are arbitrary functions, then 0x = 0 does not imply (-x)y = -xy.

The catalogue i was given is the following one:

1) x+y=y+x.....................xy = yx

2) x+(y+z)=(x+y)+z....................x(yz)=(xy)z

3) x+0 =x..................................1x =x

4) x+(-x) =0..............................$\displaystyle x\neq 0\Longrightarrow\frac{1}{x}.x=1$

5) x(y+z) =xy + xz
• Jun 4th 2012, 10:13 AM
emakarov
Re: equivalence
This is a complete set of axioms of a field. All three properties in post #1 are corollaries of field axioms, and all implications in post #3 are provable (from field axioms) without even using the assumptions (though these assumptions can be proved as intermediate results). So in this sense properties in post #1 are equivalent. However, it does not make much sense to claim such equivalence. It makes sense to claim that A and B are equivalent when the underlying theory proves neither A nor B, but together with A the theory proves B and together with B it proves A. In this sense, given axioms of real numbers except the least upper bound axiom, the latter is equivalent to Bolzano–Weierstrass theorem. All theorems of mathematics are also equivalent, but it does not make much sense to say that commutativity of addition on natural numbers is equivalent to Gödel's incompleteness theorem.

That said, here are proofs of some implications from post #3. It follows from group axioms that additive inverse is unique. Suppose that 0y = y for all y. Then xy + (-x)y = (x + (-x))y = 0y = 0, so (-x)y and -(xy) are both inverses of xy.

Also, it follows from group axioms for addition that --x = x (both are inverses of -x). Suppose (-x)y = -xy; then by commutativity x(-y) = -xy. So, (-x)(-y) = -(x(-y)) = --(xy) = xy.

I don't know a short proof of 0x = x from (-x)(-y) = xy. It seems easier to prove it directly. Namely, 0x + 0x = (0 + 0)x = 0x, so adding -(0x) on both sides gives 0x = 0.
• Jun 6th 2012, 05:14 AM
psolaki
Re: equivalence
Quote:

Originally Posted by emakarov
This is a complete set of axioms of a field. All three properties in post #1 are corollaries of field axioms, and all implications in post #3 are provable (from field axioms) without even using the assumptions (though these assumptions can be proved as intermediate results). So in this sense properties in post #1 are equivalent. However, it does not make much sense to claim such equivalence. It makes sense to claim that A and B are equivalent when the underlying theory proves neither A nor B, but together with A the theory proves B and together with B it proves A. In this sense, given axioms of real numbers except the least upper bound axiom, the latter is equivalent to Bolzano–Weierstrass theorem. All theorems of mathematics are also equivalent, but it does not make much sense to say that commutativity of addition on natural numbers is equivalent to Gödel's incompleteness theorem.

That said, here are proofs of some implications from post #3. It follows from group axioms that additive inverse is unique. Suppose that 0y = y for all y. Then xy + (-x)y = (x + (-x))y = 0y = 0, so (-x)y and -(xy) are both inverses of xy.

Also, it follows from group axioms for addition that --x = x (both are inverses of -x). Suppose (-x)y = -xy; then by commutativity x(-y) = -xy. So, (-x)(-y) = -(x(-y)) = --(xy) = xy.

I don't know a short proof of 0x = x from (-x)(-y) = xy. It seems easier to prove it directly. Namely, 0x + 0x = (0 + 0)x = 0x, so adding -(0x) on both sides gives 0x = 0.

That inspires me to ask : Is there a method of finding out whether two theorems are equivalent or not .

For example how can we show whether or not : (-x)y =-xy and $\displaystyle \frac{1}{x}.\frac{1}{y} = \frac{1}{xy}$ are equivalent
• Jun 6th 2012, 05:23 AM
emakarov
Re: equivalence
As I said, two theorems are always trivially equivalent. If A and B are theorems, say, of field axioms, then A -> B is provable just because B is provable. Inquiring about equivalence of two statements makes sense when they are not theorems.

There is a topic in mathematical logic called "reverse mathematics." From Wikipedia:
Quote:

In reverse mathematics, one starts with a framework language and a base theory—a core axiom system—that is too weak to prove most of the theorems one might be interested in, but still powerful enough to develop the definitions necessary to state these theorems. For example, to study the theorem “Every bounded sequence of real numbers has a supremum” it is necessary to use a base system which can speak of real numbers and sequences of real numbers.

For each theorem that can be stated in the base system but is not provable in the base system, the goal is to determine the particular axiom system (stronger than the base system) that is necessary to prove that theorem. To show that a system S is required to prove a theorem T, two proofs are required. The first proof shows T is provable from S; this is an ordinary mathematical proof along with a justification that it can be carried out in the system S. The second proof, known as a reversal, shows that T itself implies S; this proof is carried out in the base system. The reversal establishes that no axiom system S′ that extends the base system can be weaker than S while still proving T.
Note that the base theory is too weak to prove the axiom and the theorem in question.
• Jun 6th 2012, 06:59 AM
psolaki
Re: equivalence
Quote:

Originally Posted by emakarov
As I said, two theorems are always trivially equivalent. If A and B are theorems, say, of field axioms, then A -> B is provable just because B is provable. Inquiring about equivalence of two statements makes sense when they are not theorems.

There is a topic in mathematical logic called "reverse mathematics." From Wikipedia:Note that the base theory is too weak to prove the axiom and the theorem in question.

That would destroy the independence of the axioms
• Jun 6th 2012, 07:06 AM
emakarov
Re: equivalence
Not sure: what would destroy the independence of the axioms?