robly18 wrote:It does break an axiom because 0x = 1 is impossible.
Okay, so are you citing the Zero Product Property axiom? If so, then 0*∞, as I said equals 0, so there is no axiom breakage. Including 1 as another solution doesn't break anything. Technically, it would be incorrect to pull 1 straight out of the solutions for 0*∞ to say 0*∞=1, so it still doesn't break that axiom, but if you think my use of the word "technically" makes my statement bad (sorry for my lack of a better descriptive word), then remember that
technically, the axiom should have a special rule in the case of 1/0. I given my logic for this, if you must put your faith wholeheartedly in this axiom, then show your logic, too.
robly18 wrote:The difference here is that there is no axiom contradicting that. There is no axiom saying (x^n)^1/n equals x. You could say that then it would equal x^(n/n) but that's just a rule. No axioms actually say that: it's just a neat trick that can be used in most cases.
Hmm... It seems that you're right. I've searched it up and sources say it is only used for integer exponents. What I don't understand is, how would you ever solve quadratic, cubic, radical, etc. equations then? For example:
x^3=1
How would you solve that? By raising each side to 1/3? Well, if the power of a power property doesn't work for non-integer exponents, then you can't use it in this instance. Anyways, that wasn't the point of my comparison. My point was to show you how something could return multiple values, and how someone (you) might think that would mean it was impossible.
robly18 wrote:(x^n)^1/n equals x
No no no. There isn't a property in the world that says that (if you're using the official algebraic system that is). (x^n)^(1/n)=x, not (x^n)^1/n, and yes, it does make a difference. (Okay, maybe not).
robly18 wrote:Look, how about this: let's stop this argument
Argument? To me, it felt like I was explaining what you didn't understand. No arguing. I mean, it was getting a little back-and-forth with that axiom stuff, but I tried to solve that in my last post. Still, you insist, it must violate an axiom. Axiom. Axiom. That's actually a little fun to say. And, stop it? Why? Do you not like it?
I find it exciting to find what you have come to tell me, and how I should reply to let you see my thoughts on that matter.
robly18 wrote:until either of us has better proof to send at each other than "breaks axioms" "does not".
Please define 'better'. I can see that my current words have obviously been failing to universally show the existence of 1/0, but I don't know how to phrase it any better. I was thinking I should have just made my thoughts more concise, but I tried that last time, and I still repeat. You know, it would be helpful if you could show each of your 'disproofs' to me in one post so that I can 'dis-disprove' them specifically instead of applying the general "There are multiple solutions!" explanation. If not, that's fine. I also think that it would be useful for me to bullet-point my own arguments here:
-You can't say, for example, 1/0*0=1, because 1/0*0={all real numbers}, meaning saying it equalled one would be like saying 1^0.5 = 1, which leaves out the solution -1, which is problematic, as you have pointed out (you see, the problem isn't in 1/0, but in your assumption that 1/0*0 must equal 0 and nothing else.
-If I am saying that the current consensus in mathematics is wrong, there's no reason why axioms that violate what I'm saying can't be wrong in this instance either, using pure logic (i.e. 0*x must equal 0 because... Therefore, your logic is wrong because...).
-I know there are more, but this is basically what the conversation has come down to.
robly18 wrote:So here's a homework assignment:
Define addition, subtraction, multiplication and division by a + b/0. That's what I have for you.
Easy. Given:
a and b are both nonzero real numbers.
1/0 is ∞.
∞+a=∞.
0*b=0
b/0=b/(0*b)=b/b/0=(b/b)/0=1/0
b/0=1/0, so a+b/0=a+1/0=a+∞=∞=1/0
So, a+b/0=1/0. Then you just multiply and divide and add and subtract like normal, keeping in mind that there may be more than one returned value for even algebraic operations. For example, 1/0*0=1/0*(0*a)=1/0*0*a=1*a=a, so 1/0*0 has an infinite number of possible values.
Nobody ever notices my signature. ):