Zero to the Zero

[Dapthar]

0/0=R and all imaginary numbers.

False. It is undefined. Of course, you are free to devise your own number system in which this statement is true, however, you must re-derive any other mathematical results that you wish to use. You can't change a definition, and apply other results in Mathematics, as if nothing else had changed.

if a/b=c, and a is 0, then c is 0 and can't be any other number but 0.

Yes, for any given [math]b \ne 0[/math], the above statement is true, which proves that [math]0/x = 0[/math], for non-zero real [math]x[/math].

+or-, thanks for the reminder. For the sake of logical thought and processing, since there can be two values for a quadratic equation, and 3 for a cubic, and so on, for x/0, or rather 0/0 there are in infinite number of solutions.

Nope. Division of two real numbers is a single-valued operation, as I mentioned in my earlier post.

0/0=? 1

0/0=1/1*0/0

0/0=0/0 correct

Nope. Again' date=' division by [math']0[/math] is undefined so one cannot use algebraic operations on any equation that has such a term in it.

All real numbers and imaginary numbers follow the same pattern if 1 is replaced by x or any variable that you choose.

Nope. Your argument isn't even true for real numbers, thus one cannot extend a fallacious argument to the complex domain.

It's not only undefined, it's infinity.

You're kidding, right? Do you realize that, in only 6 words, you've managed to contradict yourself?

 

If something is undefined then it cannot be [math]\infty[/math] because it is undefined.

[Dave]

It's not only undefined, it's infinity.

 

I was going to post something about this, but Dapthar beat me to it.

 

Please read the post above carefully, you really don't want to go around saying that

[eighth man]

You like big numbers?

 

Look at my thread on the universe is

a number. Actually it's not a very

big number, only to us humans.

 

All conceivable universes are

10^10^10^3.

[bloodhound]

thats not that big a number to be honest

[123rock]

I said square root, as in square root of 49 is + or - 7, since both can be solutions.

This comes down to my disagreement with the square root of an integer bein only the positive root, and the only way the negative one is a solution is if the original equation was x^2=49, which is utter stupidity.

And no, I do not make my own rules, as 0/0 has any solution, since by the definitions themselves x/y means how many times y can be contained in x, and 0 can be contained in itself 1,2,3....R numbers.

 

0/0=x, where x=x

 

0/0-x=0

0/0(0/0-x)=0/0(0)

0/0-0/0=0/0

 

Since virtually any number can be replaced by x, R or i, then x is R and i.

[123rock]

Actually if x=0/0, then...

since 0/0=x

 

then x-x=0, but the only way x-x=0 is if x isn't 0/0, or y/0, which defies the original

given that x=0/0

 

so the only way x=0/0 is if x=/ 0/0

 

hence the contradiction and the proof that x/0 is undefined, so I guess I was wrong...

[123rock]

False. It is undefined. Of course' date=' you are free to devise your own number system in which this statement is true, however, you must re-derive any other mathematical results that you wish to use. You can't change a definition, and apply other results in Mathematics, as if nothing else had changed.

Yes, for any given [math']b \ne 0[/math], the above statement is true, which proves that [math]0/x = 0[/math], for non-zero real [math]x[/math].

Nope. Division of two real numbers is a single-valued operation, as I mentioned in my earlier post.Nope. Again, division by [math]0[/math] is undefined so one cannot use algebraic operations on any equation that has such a term in it.

Nope. Your argument isn't even true for real numbers, thus one cannot extend a fallacious argument to the complex domain.

You're kidding, right? Do you realize that, in only 6 words, you've managed to contradict yourself?

 

 

If something is undefined then it cannot be [math]\infty[/math] because it is undefined.

 

I just realize I was contradicting myself, but tautology isn't really helpful, if all you're going to say is that 0/0 is undefined and so you can't use algebraic or arithmetic equations to prove it's undefined. That contradicts itself.

[Dave]

thats not that big a number to be honest

 

Well, it's a bigger number than an estimate of all particles in the known universe, so I think that accounts for being "quite big"

[123rock]

Actually if x=0/0' date=' then...

since 0/0=x

 

then x-x=0, but the only way x-x=0 is if x isn't 0/0, or y/0, which defies the original

given that x=0/0

 

so the only way x=0/0 is if x=/ 0/0

 

hence the contradiction and the proof that x/0 is undefined, so I guess I was wrong... [/quote']

 

Further extending this, we can see that 0/0 cannot be 0 either, since if it could be 0, then x can be any number which rebrings the contradiction.

 

If 0/0=0, then when we have x-x=0, 0/0-x=x-0/0, since each can be replaced with each other. If we are to apply for x=some number, for e.g. 4, then we get that 0-4=4-0, showing that -4=4, which is a contradiction. This can only lead us to suggest that x can only be 0, in which is the only case that 0/0=0 works, but since x is a variable and can be any number R, or i, then 0/0=0 is a false statement as well as 0/0=x.QED

The only thing that cannot be disproved is if 0/0=infinity.