Prove that product of two negitives equals a positive

[alinta]

I don't understand why everyone in this thread is getting upset about the lack of axiom lists instead of actually googling axioms for real numbers and then getting on with the proof. Pretty sure there's only one list of axioms for real numbers.

ANYWAY.

The following proof is to show that the product of two negitive numbers equals a positive. I will assume a(-b)=0 but this can be deduced from the axioms anyway.

a+(-a)=0
(a+(-a))x-b= 0 x -b
=0
we know ax-b=-ab

-ab+(-ax-b)=0

so -ax-b=ab

I have gotten this far but am a little stuck with showing x^2>0

my attempt looks like this:

let x=-a

from the above we know -ax-a=axa=a^2

for the case x=a

axa=a^2

however this seems a little too simple to me.

If anyone could help constructively that would be great.

 

[daon2]

I don't understand why everyone in this thread is getting upset about the lack of axiom lists instead of actually googling axioms for real numbers and then getting on with the proof. Pretty sure there's only one list of axioms for real numbers.

ANYWAY.

No, there isn't one set of axioms and there are several different constructions of the real numbers. Sometimes they are also stated differently. A beginning proofs class is very particular about what one can assume when writing a proof. Logical deductions must be very clear, and very obvious. Even if something is intuitively sound, skipping any steps might result in an "incorrect" answer.

Second, we need to know what other theorems or propositions have been proven, because always resorting to axioms would result in extremely long, extremely complicated proofs.

The following proof is to show that the product of two negitive numbers equals a positive. I will assume a(-b)=0 but this can be deduced from the axioms anyway.

Why are you assuming that? That cannot be deduced from the axioms because it is not true. Attention to rigor is the point of a class like this.

a+(-a)=0
(a+(-a))x-b= 0 x -b
=0
we know ax-b=-ab

-ab+(-ax-b)=0

so -ax-b=ab

Please use words to describe what you are trying to accomplish, I cannot make sense of what you are doing.

I have gotten this far but am a little stuck with showing x^2>0

my attempt looks like this:

let x=-a

from the above we know -ax-a=axa=a^2

for the case x=a

axa=a^2

however this seems a little too simple to me.

If anyone could help constructively that would be great.

Again, I am not sure what you are doing. Your conclusion is not true, since if x=-a, -a^3 = axa = a^2 is not true.

You need to post your axiom list and other known truths. For example, if you know that (-1)(-1) = 1, then it makes this proof much easier.

 

[HallsofIvy]

I don't understand why everyone in this thread is getting upset about the lack of axiom lists instead of actually googling axioms for real numbers and then getting on with the proof. Pretty sure there's only one list of axioms for real numbers.

As Daon2 said, no that is not the case.

ANYWAY.

The following proof is to show that the product of two negitive numbers equals a positive. I will assume a(-b)=0 but this can be deduced from the axioms anyway.

a+(-a)=0
(a+(-a))x-b= 0 x -b
=0
we know ax-b=-ab

-ab+(-ax-b)=0

so -ax-b=ab

I have gotten this far but am a little stuck with showing x^2>0

Up to this point, you appear to have been using "x" as the multiplication symbol. Now, suddenly, it is a variable. That is very confusing!

my attempt looks like this:

let x=-a

from the above we know -ax-a=axa=a^2

So you have (-a)^3d??? No, having said "x= -a" you have reverted to "x" as multiplication symbol!

for the case x=a

axa=a^2

however this seems a little too simple to me.

If anyone could help constructively that would be great.

 

[lookagain]

I don't understand why everyone in this thread is getting upset about the lack of axiom lists instead of actually googling axioms for real numbers and then getting on with the proof. Pretty sure there's only one list of axioms for real numbers.

As Daon2 said, no that is not the case.


Up to this point, you appear to have been using "x" as the multiplication symbol. Now, suddenly, it is a variable. That is very confusing! alinta, not only does it appear you have been using "x" also for multiplication, you definitely have. Not only is it very confusing, it is wrong. Where needed/appropriate, use grouping symbols and/or an asterisk for multiplication among some combination of constants and variables.


So you have (-a)^3d??? No, having said "x= -a" you have reverted to "x" as multiplication symbol!

.