The answers to 1^(1/2) and 1/0

[testtubegames]

I see your point, and exfretinfinity was what I was using the symbol for before, but I'm still using the same exact operations. ... I'll define the number , such that + 1 =

I see a problem already. If that's regular addition you're using, then it is easy to show that does not exist. (Subtract from both sides... 0=1...oops! does not exist) *Or* you're using exfretAddition... where " exfretPlus x = " is one of the core axioms of your operation.

My point is you're sunk at that first step. *Unless* you take this outside the ruleset of mathematics and invent your own mathematics. Which is fine and fun, but, again, doesn't show 1/0=infinity... but rather 1 exfretDivided by 0 =

[exfret]

I see your point, and exfretinfinity was what I was using the symbol for before, but I'm still using the same exact operations. ... I'll define the number , such that + 1 =

I see a problem already. If that's regular addition you're using, then it is easy to show that does not exist. (Subtract from both sides... 0=1...oops! does not exist) *Or* you're using exfretAddition... where " exfretPlus x = " is one of the core axioms of your operation.

My point is you're sunk at that first step. *Unless* you take this outside the ruleset of mathematics and invent your own mathematics. Which is fine and fun, but, again, doesn't show 1/0=infinity... but rather 1 exfretDivided by 0 =

I see your concern, and I have actually explained this before:

Man, you sure have great skill at making walls of text my brain hurts when reading.
As for the not a number thing, here's why it isn't a number:
The axioms apply for everything. They're the basis of mathematics. They are literally what defines our system of counting, adding, subtracting... If something doesn't obey these axioms, then it isn't really a thing.
As for the "how can it not be a number thing" let me make an analogy.

Say, n = n + 1. Is this possible?
As my good old friend Jim would say, this would mean 0 = 1. It's not possible as you can see.
Jim isn't a real person. Anymore, at least.

Anyway, as for the n = 1/0, here's why it's wrong:
Multiply both sides by zero
0n = 1
Use axiom proofs to figure out this:
0n = 0
And then ,you get 0 = 1.
This is... not really possible.
Therefore, the existence of a number that is equal to one over zero is impossible, just like the existence of a number equal to itself plus one. They do not obey the axioms and cause inconsistencies. It may sound like a case of "it doesn't obey what we know, so sweep it under the rug", but that's the thing. Their existence would make mathematics inconsistent. We call the undefined because, with our set of axioms, they cannot be defined.

Actually, n=n+1->0=1 is not exactly correct. You see, when you write the whole thing out, something pops up:

n=n+1
-n -n
n-n=n-n+1

Now, you might say: "Hey! n-n=0, right?" Well, that's right, but remember, we're taking an operation here, and operations can return multiple values. This may hurt your head, but it happens a lot when it comes to ∞, and it happens all the time with square roots, too, so it isn't an entirely new concept. Now, remember, since n=n+1, we can plug in n+1 in as n, and we get this:

n-n=n-(n+1)+1
0=n-n-1+1
0=0+0
0=0

Now, before you start telling me how this violates the basic laws of mathematics, I never said n-n wasn't equal to 0. It just could be equal to other things, too. This doesn't mean 0 isn't one of its 'solutions'. You have to remember that some operations have multiple values, but only one of them is the 'right' one (remember that discussion about 0/0?), and sometimes we can even deduce which one is right, but you can't just pick out one of those values and plug it in, because it'll get contradictory things, like you've shown. In this case, n-n actually can't be 0 for both sides of the equation. It's kind of like derivatives. A derivative is actually just 0/0, but by doing some cool tricks, you can find out which of the values returned by the operation 0/0 was right. n-n is the same way: it's indeterminate, and that means you can't just assume one of it's values is correct.

Adding on to what I wrote before, if I said there was a number, i, such that sqrt(-1) = i, then you could say that would be impossible, because when you raise each side of the equation to the power of 4, you get i^4=1, and then when you take the fourth root of that, you get i=1, which obviously isn't correct, because 1^2 ≠ -1. Well, it certainly seems like we haven't had this problem before, right? viewtopic.php?f=1&t=62. Right?

Ahh, the conversation's gotten to the point where I can just copy and paste text. Such an easier life...

Anyways, the fact that we're not used to multiple values being returned from an operation just makes this all seem controversial. It's probably 80% of y'all's concerns. Oh, robly, now do you see how not teaching multiple returned values from an operation early on can have an impact? Most of this is so counter-intuitive to the average person now... no wonder it's been called undefined until now!

P. S. I still agree with you: it would probably only confuse people more, but it's a good reason for why I might be the first person (I know of) to have thought of this.

[robly18]

Look, here's the thing:
Our axioms do not support this.
Our axioms do not support a number that, when multiplied by zero, is not zero.
Our axioms explicitly say that, for any number a in R (and also C), 0a = 0. No exceptions.
You could create a new set of axioms. You could say 1/0 is not in R, but in D instead. But that would be creating a new set. And our axioms for our current sets simply do not endorse this.
I mean, I myself once taught about this. The same way sqrt(-1) = i, I thought of defining 1/0 as d (for division). I then proceeded to see if the set D, with d and its multiples, proved consistent. I remember not doing much work, but y'know what: let's define it right now.
Take a number in D. It has two parts: the real part and the... divided part?
So, a number in D is represented as this: (a, b).
So 1 would be represented as (1, 0)
1/0 would be represented as (0, 1)
So a + b/0 is represented as (a, b).
Let's create a set of axioms based on what we know and see if it's consistent:
0(a+b/0) = b. So there is a number (0, 0) so that (a, b)(0,0) = (b, 0)
We could also define addition. That's an easy one:
(a,b) + (c,d) = (a+c, b+d)
Multiplication? Let's give it a shot:
(a,b)(c,d) = ?
Let's try it out with a simple example.
(1,0)(1,0) = (1,0).
Fair enough. How about this:
(0,1)(0,1) = (0,1)
But what happens when we mix both?
(1, 0)(0,1) = (0,1)
Multiplication by one, not very impressive. Let's try something else:
(1,1)(0,1)
This is harder. Let's put it into an easier way to recognize it:
(1 + 1/0)(1/0)
This equals 1/'0 + 1/0 = 2/0
So (1,1)(0,1) = (0,2)
Let's try this universally now. We can handle it, right?
(a,b)(c,d)
This is:
(a + b/0)(c + d/0) = a(c+d/0) + b(c/0 + d/0) = ac + ad/0 + bc/0 + bd/0 = (ac, ad+bc+bd)
Alright, so we have formally defined multiplication in D as (a,b)(c,d) = (ac,ad+bc+bd)
What happens if we do division? Let's try doing the inverse of (a,b)
1/(a+b/0)
So we need to make this so that there all terms in the lower side of the fraction are either divided by zero or not. Let's try plugging in some numbers: that might help. Let's go with good old (1,0)
1/(1 + 0/0) = 1
Okay. How about (0,1)
1/(0+ 1/0) = 0/1 = 0
Maybe we could generalize it:
1/(a+b/0)

I, myself am unable to do it at this point. Mostly because it's midnight and I should probably be headed to bed.
However, if you want to be sure that your new system is foolproof, you should keep making these formal definitions. This is how complex numbers were defined.
Please, build upon my work. Make division, and put it to the test.

[exfret]

Look, here's the thing:
Our axioms do not support this.

Except they do.

Our axioms do not support a number that, when multiplied by zero, is not zero.

When you multiply my number by zero, zero is in fact one of the numbers returned from that operation. (Look at it this way: it's indeterminate).

Our axioms explicitly say that, for any number a in R (and also C), 0a = 0. No exceptions.

Solution: How about we call my number an explex number, which doesn't belong to R or C.

Sorry, didn't have time to reply to the rest of your words. Late 4 me 2.

[robly18]

Look, here's the thing:
Our axioms do not support this.

Except they do.

Our axioms do not support a number that, when multiplied by zero, is not zero.

When you multiply my number by zero, zero is in fact one of the numbers returned from that operation. (Look at it this way: it's indeterminate).

Fine, let's play your game. Let's say there is a number in R which is equal to one over zero.
Now what?
We could divide it by zero. So we'd get:
1/0 / 0 = 1/0
Okay... Wait what? Let's do that again with algebra.
x/0 = x
Multiply both sides by zero...
x = 0x
Now, if you saw this equation, out of context. What would be the answer? I can already tell you it's 0.
Just for the kicks and giggles, let's solve it.
Our friend axiom says that for every number in C, there is a number 0 which 0x = 0
Thus, we conclude, x = 0
Thus, we just got to the conclusion 1/0 = 0

And in case you come at me with the argument that I forgot a step, here it is, and its consequences:
0x/0 = 0x
x* 0/0 = 0x

Now, we could say 0/0 is equal to zero. After all, the equation 0 = 0x works when you replace x by zero. So let's give it a shot.
0x = 0x
Sure, this applies truth. However, does the equation work when it's replaced by 1?
0 = 0*1
Yes, it does.
So we get 1*x = 0x = 0 because axioms.

Look, dividing by zero is a whole can of worms. Axiomatic worms.

Pro tip: don't.

Also, side note, please stop using the higgs symbol so often. I know I started it (guilty as charged) but in this kind of mathematical texts it just makes it hard to read.

[exfret]

Sorry for taking so long to reply. Also sorry that I don't have much time to reply, so just say and I'll create a reply to your desired level of detail (aka, say, for example, reply to every few sentences, or maybe if you're feeling particularly evil, every pixel of the post). Anyways, thanks for the reply but something I've been noticing is that you're sticking strongly with counter proofs, and the reason I haven't been accepting them is:

Multiple values returned from an expression (like in the case of 1^(1/2)).

Basically, you're saying that 1/0 either violates an axiom or is contradicting (I believe you went with a mix of the two this time, because you used an axiom in your proof). What I'm trying to say is that when you do things like 1/0*0, it doesn't just return 1, or just 0, but rather both at the same time, meaning it violates no axioms, and I am also stressing that even if it did, your counter-proof wouldn't be valid if the logic those axioms were founded upon failed in that instance. How about I tell a story this time:

Mel says: "You know squaring numbers is impossible."
Rex F. T. corrects him by saying that it is, in fact possible.
"But then when you raise to a half, you should get the number that you squared, right?"
"Well, yes."
"But 1 to the power of a half is either one or negative one, and if it's one, negative one squared gets one, but raising to the power of a half doesn't get you negative one, and if one to the half is negative one, then raising one squared to one half gets you negative one, even though it should return one. Therefore, a system where you can square numbers contradicts itself, and it violates the basic algebraic property that a number to the power of 2 then raised to the power of 1/n is said number."
To this, Rex replies: "Ah, but remember, you can't just assume one value is correct. One to a half will always be either negative or positive one. You can't just pick one out on the fly. With context, though, you may be able to pick out the right value."

[robly18]

Sorry for taking so long to reply. Also sorry that I don't have much time to reply, so just say and I'll create a reply to your desired level of detail (aka, say, for example, reply to every few sentences, or maybe if you're feeling particularly evil, every pixel of the post). Anyways, thanks for the reply but something I've been noticing is that you're sticking strongly with counter proofs, and the reason I haven't been accepting them is:

Multiple values returned from an expression (like in the case of 1^(1/2)).

Basically, you're saying that 1/0 either violates an axiom or is contradicting (I believe you went with a mix of the two this time, because you used an axiom in your proof). What I'm trying to say is that when you do things like 1/0*0, it doesn't just return 1, or just 0, but rather both at the same time, meaning it violates no axioms, and I am also stressing that even if it did, your counter-proof wouldn't be valid if the logic those axioms were founded upon failed in that instance. How about I tell a story this time:

It does break an axiom because 0x = 1 is impossible.

Mel says: "You know squaring numbers is impossible."
Rex F. T. corrects him by saying that it is, in fact possible.
"But then when you raise to a half, you should get the number that you squared, right?"
"Well, yes."
"But 1 to the power of a half is either one or negative one, and if it's one, negative one squared gets one, but raising to the power of a half doesn't get you negative one, and if one to the half is negative one, then raising one squared to one half gets you negative one, even though it should return one. Therefore, a system where you can square numbers contradicts itself, and it violates the basic algebraic property that a number to the power of 2 then raised to the power of 1/n is said number."
To this, Rex replies: "Ah, but remember, you can't just assume one value is correct. One to a half will always be either negative or positive one. You can't just pick one out on the fly. With context, though, you may be able to pick out the right value."

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.

Look, how about this: let's stop this argument until either of us has better proof to send at each other than "breaks axioms" "does not". So here's a homework assignment:
Define addition, subtraction, multiplication and division by a + b/0. That's what I have for you.

[exfret]

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.

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.

(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).

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.

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.

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.

[exfret]

1 week and... Oh who cares? You all get the point.

[robly18]

I figured out something wrong.
The concept of infinity.
I gave it some thought and found out a mistake: defining infinity as a number greater than all others.
Take +infinity + 1. If +infinity > anything that isn't +infinity, then +infinity > +infinity + 1. Subtract +infinity from both sides and you get 0 > 1, a false statement.
Now, say you make it +infinity + 1 = +infinity. That being the case, then +infinity > +infinity. Subtract infinity from both sides and you get 0 > 0
Infinity is a concept, not a number. It doesn't work as a number. The end.