After 6 years I still don't understand this

After 6 years I still don't understand this - Printable Version

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After 6 years I still don't understand this - butterfλi - 2011-03-14

Devil Wrote:X^0=1 is only so because scientists across the world agreed that they will call it that way. It could be 0 too, but since that is not possible for humans to calculate at this moment (same way as dividing something by 0 is "impossible"), we all just agree on that is must be 1 to keep things simple.

It's the same as the time-travel paradox, it shouldn't be possible, that's why scientists agreed it isn't possible.

Math isn't about truth, it's about logic. Logic that isn't possible isn't logic, it's chaos. Although chaos is possible, we all agree that it isn't, just to keep things simple...

Uhmm... no.

Math, like any science, is formulated through hypothesizing, experimenting and concluding. It's just less observation based like say astronomy. Theorems in math are tested theories and hold true for as long as until a new theory disproves it. In other sciences, theories are disproved by a newfound observation, which does happen every now and then. In math, theorems are logically proven so you do not see a theorem disproved very often (or at all) unless there is a logical fallacy. And like science, new theories and concept such as calculus and the imaginary number revolutionize and advance the field.

Division by zero isn't because people "agreed it's impossible". It is possible. It's just that it yields a logical error (1 = 0 or 1 = 2). 1 + 1 = 2 isn't because people agree. There is a theorem that proves that 1 + 1 = 2 is in fact true and it doesn't involve counting fingers and a bunch of scientists nodding their heads in agreement.

" Wink

After 6 years I still don't understand this - Devil - 2011-03-14

butterfλi Wrote:Uhmm... no.

Math, like any science, is formulated through hypothesizing, experimenting and concluding. It's just less observation based like say astronomy. Theorems in math are tested theories and hold true for as long as until a new theory disproves it. In other sciences, theories are disproved by a newfound observation, which does happen every now and then. In math, theorems are logically proven so you do not see a theorem disproved very often (or at all) unless there is a logical fallacy. And like science, new theories and concept such as calculus and the imaginary number revolutionize and advance the field.

Division by zero isn't because people "agreed it's impossible". It is possible. It's just that it yields a logical error (1 = 0 or 1 = 2). 1 + 1 = 2 isn't because people agree. There is a theorem that proves that 1 + 1 = 2 is in fact true and it doesn't involve counting fingers and a bunch of scientists nodding their heads in agreement.

" "

Math = number of rules agreed by scientists all over the world with hypothesizing how logic should be written down.

It's only rules, not truth. That's why they invented 0 and infinity, just to keep the logic going.

That dividing by zero leads to a logical arrow, is the result of the rules of logic.

Wink

After 6 years I still don't understand this - shouri - 2011-03-14

We can cut the philosophical discussion regarding the nature of logic at this point. This is just like the threads involving why .999999...=1. People will just argue back and forth and nothing's going to get accomplished. 0 ^ 0 isn't defined as zero nor 1. It's indeterminant. There's no question about that. Some want it to be zero, some want it to be 1. It's neither one at the moment. It's like asking what 1/0 is.

OP asked a question, we've answered it hopefully.... let's not continue to derail.

After 6 years I still don't understand this - Noah - 2011-03-14

XTOTHEL Wrote:x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

That is actually the proof that x^0 = 1 for all nonzero x. Well, apart from saying that y = z - z.

shouri Wrote:0 ^ 0 isn't defined as zero nor 1. It's indeterminant. There's no question about that. Some want it to be zero, some want it to be 1. It's neither one at the moment. It's like asking what 1/0 is.

Well, actually, there are several issues by saying that it is neither one, as it would make more sense to say that it is both, mainly depending on whether you work with discrete or continuous values. I think it is a funny think to play around with, though. First of all, we know from calculus that

[Image: 5rsquj5.png]

- right? So, set n = 1:

[Image: 48zm5vk.png]

So, the derivative of x is usually considered to be equal to 1 for all values of x. In order to make that happen, x^0 needs to be 1.

However, according to limits, we know that

[Image: 6l65yu7.png]

and that

so by this step, we know that it is undetermined, and we can stop the non-believers into thinking it is something else by proof! Big Grin

Noah

After 6 years I still don't understand this - shouri - 2011-03-14

Noah Wrote:That is actually the proof that x^0 = 1 for all nonzero x. Well, apart from saying that y = z - z.

Well, actually, there are several issues by saying that it is neither one, as it would make more sense to say that it is both, mainly depending on whether you work with discrete or continuous values. I think it is a funny think to play around with, though. First of all, we know from calculus that

- right? So, set n = 1:

So, the derivative of x is usually considered to be equal to 1 for all values of x. In order to make that happen, x^0 needs to be 1.

However, according to limits, we know that

and that

so by this step, we know that it is undetermined, and we can stop the non-believers into thinking it is something else by proof!

Noah

well, what I meant by neither, is that someone is not allowed to say "it's definitely 1" or "it's definitely 0". Bad wording on my part.

Also, using limits says nothing about the actual value at 0. So the limit way doesn't help. We at least get to show that those functions tend to different values, but those functions aren't continuous at x=0... so the two limits you showed don't prove that it's undetermined... technically. Guess it's a nice way of trying to make someone realize that that is the case though.

laters~

Yay for noah Big Grin

After 6 years I still don't understand this - GummyBear - 2011-03-14

The answer to 10^0 = 1 is the same answer as 2 > 1. That answer is quite simple really, it's 42.

After 6 years I still don't understand this - octopusprime - 2011-03-15

shouri Wrote:The 0^0 argument is like the why .999... = 1 argument (sort of).

Not really. .999..... = 1 because there is no number between them therefore they are the same number. It's literally the same reason why 1 = 1.

Anyone who argues against this just doesn't get math's.

After 6 years I still don't understand this - shouri - 2011-03-15

octopusprime Wrote:Not really. .999..... = 1 because there is no number between them therefore they are the same number. It's literally the same reason why 1 = 1.

Anyone who argues against this just doesn't get math's.

^obviously didn't read the rest of my post. I know this fact very well. I'm saying that this situation is similar in that it leads to pointless arguing without anyone learning anything.

After 6 years I still don't understand this - Locked - 2011-03-15

Ahaha

I should really put /sarcasm tags at the end of my posts. 0^0 was just being silly.

octopusprime Wrote:Not really. .999..... = 1 because there is no number between them therefore they are the same number. It's literally the same reason why 1 = 1.

Anyone who argues against this just doesn't get math's.

No.

After 6 years I still don't understand this - octopusprime - 2011-03-15

Locked Wrote:No.

yes.

After 6 years I still don't understand this - Bloodjedi - 2011-03-15

Such a bad idea to look into this thread LOL

After 6 years I still don't understand this - 2147483647 - 2011-03-15

Noah Wrote:

If you approach this from 0-, you get 0/0 (indeterminate). Only if you approach it from 0+ do you get 0. Therefore, because there is a discontinuity here, it would make more sense to define 0^0 using the first limit you provided,

Noah Wrote:

since it is continuous at all points except for that one hole at 0. In other words, both x^-0 is 1 and x^+0 is 1, whereas only 0^+x is 0. For a limit to be defined at a single point, both sides of the limit must be satisfied.

Otherwise, the step function f(x) = floor(x) would equal both 1 and 0 at x=1, when it can only actually exist at one of the two points or the function would not be a function.

Furthermore:

k^x = y
x*ln(k) = ln(y)
ln(k) = ln(y)/x
k = e^(ln(y)/x)
k = y^(1/x)

~~If x = 0, there is just no possible selection of y that can make k = 0.~~ poor wording

After 6 years I still don't understand this - Noah - 2011-03-15

shouri Wrote:Also, using limits says nothing about the actual value at 0. So the limit way doesn't help. We at least get to show that those functions tend to different values, but those functions aren't continuous at x=0... so the two limits you showed don't prove that it's undetermined... technically. Guess it's a nice way of trying to make someone realize that that is the case though.

Oops, can't prove it, but can at least show that there is some weird things going around!

2147483647 Wrote:If you approach this from 0-, you get 0/0 (indeterminate). Only if you approach it from 0+ do you get 0. Therefore, because there is a discontinuity here, it would make more sense to define 0^0 using the first limit you provided,

It's a very good idea to try and approach this from 0- to check out what the actual value is, but that's where we head into some problems: 0^x for negative values of x is not defined (0/0 for all negative values). So we can't really say that 0^x is not defined for x = 0, though that would have been a really good idea. It's not a discontinuity per say, because it's just a one-sided test, and we can't test it from the other side.

2147483647 Wrote:it would make more sense to define 0^0 using the first limit you provided, since it is continuous at all points except for that one hole at 0. In other words, both x^-0 is 1 and x^+0 is 1, whereas only 0^+x is 0. For a limit to be defined at a single point, both sides of the limit must be satisfied.

The main idea behind the choosing of those limits is just to show that for some limits, it goes to zero, as for others, it goes toward one.

Spoiler

2147483647 Wrote:Otherwise, the step function f(x) = floor(x) would equal both 1 and 0 at x=1, when it can only actually exist at one of the two points or the function would not be a function.

True enough. What we need to find though, is the actual value of 0^0. If we knew that answer as a rational number, it would be easy to find floor(0^0).

2147483647 Wrote:Furthermore:

If x = 0, there is just no possible selection of y that can make k = 0.

Another great example that 0^0 is a weird thing!

This Schrödinger-number is funny, as we cannot look inside the theoretical box. Sad

Noah

After 6 years I still don't understand this - 2147483647 - 2011-03-15

Actually, what I meant was that if 0^0=0, then x^k = y must look exactly like y^(1/x) = k. After all, 0^0=0 should still be 0^0=0 when rearranged around.

In this case, k = 1/x. Notice that 1/x can never finitely be 0. In fact, 1/x diverges around the point 0. The only way to make k or x equal to zero is to make the other (+/-) infinity, which violates the initial condition that k=x=0.

I have yet to see a complete proof that 0^0 can ever equal 0. There's always a discontinuity there. People who claim that 0^0=0 only take one-sided limits, which is just isn't as strong as the numerous two-sided limit proofs that exist for 0^0=1. Actually, I would be surprised to see an everywhere continuous (except at 0) function that makes p^q=0 when approached from both sides of the point, (0,0).

After 6 years I still don't understand this - thinbear - 2011-03-16

trying to explain it in real life example:

given an empty box of nothing but bacteria with population of 1, it double its population every 1 hour;
so 5 hours later it would be 2^5 = 32
10 hours later will be 2^10 = 1024

what happen at hour 0? it won't be 0 bacteria, with 0 bacteria there will be nothing going on

After 6 years I still don't understand this - 2147483647 - 2011-03-16

What you're saying is that the rate of change of the bacteria is proportional to its current population (that's how you get exponential growth). Therefore, you claim:

dx/dt = kx

dx/dt -kx = 0

(dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) = 0

∫ (dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) dt = C

x*(e^(-kt)) = C

x = C*e^(kt)

x = C*(e^k)^t

Since you specified that the bacteria is doubling in population every unit of time, e^k = 2. Therefore, you have:

x = C*2^t

You also specified that the initial condition that there is one unit of bacterial population. Therefore:

1 = C*2^(0) = C

C = 1

Thus, your model states:

x = 1*2^t

If your initial condition is 0, then:

x = C*2^t

0 = C*2^(0) = C

C = 0

x = 0*2^t = 0

Obviously, in an initial population of 0, there will be nothing going on, because doubling 0 will result in 0 and so on. You haven't proven anything at all regarding 0^0.

shouri Wrote:The 0^0 argument is like the why .999... = 1 argument (sort of). In that people tend to argue craploads for multiple sides and no one like to agree with the other side. 0^x = 0 for any nonzero, thus people say that 0 ^ 0 should be 0. But x ^ 0 = 1 for any non zero x, thus people say that 0^0 should be 1.

Actually, the .999... = 1 argument is completely different from the 0^0 argument. By using summations, fractions, and various other methods, there actually exists valid proof that .999... = 1. For 0^0=0, all arguments supporting 0^0=0 are incomplete, including the 0^x argument.

After 6 years I still don't understand this - shouri - 2011-03-16

2147483647 Wrote:Actually, the .999... = 1 argument is completely different from the 0^0 argument. By using summations, fractions, and various other methods, there actually exists valid proof that .999... = 1. For 0^0=0, all arguments supporting 0^0=0 are incomplete, including the 0^x argument.

Oh my god... do I HAVE to say it again? I'm NOT saying that these two arguments are the same MATHEMATICALLY. I'm saying that we're running into the SAME problem: people arguing back and forth without anything being accomplished because the people who believe one side of the argument will never be convinced by the other.

After 6 years I still don't understand this - 2147483647 - 2011-03-16

shouri Wrote:Oh my god... do I HAVE to say it again? I'm NOT saying that these two arguments are the same MATHEMATICALLY. I'm saying that we're running into the SAME problem: people arguing back and forth without anything being accomplished because the people who believe one side of the argument will never be convinced by the other.

Can you read? I'm saying that your statement (bolded) would be true if there had existed valid proofs supporting both sides of the argument. In this case, there is only valid proof supporting one side in the 0^0 problem. It's not the same mathematically, logically, or any other way. Tongue

Unless you're saying that mathematics is actually religion, which would be absolutely moronic.

After 6 years I still don't understand this - Noah - 2011-03-16

2147483647 Wrote:Can you read? I'm saying that your statement (bolded) would be true if there had existed valid proofs supporting both sides of the argument. In this case, there is only valid proof supporting one side in the 0^0 problem.

Hmm, have I missed something? As far as I know, there are no proofs that says 0^0 = 1 or 0^0 = 0.

Noah

After 6 years I still don't understand this - 2147483647 - 2011-03-16

Noah Wrote:Hmm, have I missed something? As far as I know, there are no proofs that says 0^0 = 1 or 0^0 = 0.

Noah

You stated a few yourself. I'll even add in a bonus. Here is the plot of f(x,y) = x^y

[Image: msp526919egh73abi38f26g.gif]

The point (0,0) is clearly on 1.