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Author Topic: m/n where n=0  (Read 8385 times)

rutsber

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m/n where n=0
« on: August 15, 2011, 02:09:37 am »

So, I just got off a high from winning an argument about dividing by zero, and I'm kind of wondering if there is any particular reason why you can't divide by zero. As far as I know, it is simply undefined, but no particular law of mathematics states that you can't. If anyone who knows more about math than me (not hard) knows of a law that specifies that you cannot, please tell me what the law is, for future reference.
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G-Flex

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Re: m/n where n=0
« Reply #1 on: August 15, 2011, 02:16:26 am »

As far as I know, it is simply undefined, but no particular law of mathematics states that you can't.

It's undefined because it's meaningless and breaks shit. In fact, a lot of "troll math" proofs rely on covertly getting you to divide by zero and pretend the results are valid. Basically, you can get division by zero to break plenty of established rules.

If you want to know more, look up the topic on Wikipedia, as well as "indeterminate forms". As a rational expression gets closer and closer to dividing by zero, it doesn't necessarily approach any particular value even if the numerator is approaching the same number each time. If you have (x/y) with y approaching zero and x approaching 7, you get different results depending on which is approaching its value from which direction (negative or positive). In a context like this, for instance, there's no good way to define the limit.
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Kay12

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Re: m/n where n=0
« Reply #2 on: August 15, 2011, 02:19:39 am »

It depends on setting. If you do that in normal arithmetic, the result is undefined. Whether you consider it a "law" or not that you're pretty limited when dealing with "undefined" numbers is not up to me, of course.

Computers, on the other hand, may have exceptions when dividing by zero. Again, the reason is that you can't really compute with undefined numbers. Some cases will return infinity (positive or negative) or a special value (NaN) but each of these works pretty badly in calculations if they weren't expected.

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G-Flex

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Re: m/n where n=0
« Reply #3 on: August 15, 2011, 02:21:54 am »

Computers, on the other hand, may have exceptions when dividing by zero. Again, the reason is that you can't really compute with undefined numbers. Some cases will return infinity (positive or negative) or a special value (NaN) but each of these works pretty badly in calculations if they weren't expected.

It's important to note here that these are just industry standard methods for dealing with division by zero, and do not necessarily reflect any mathematical fact.
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Kay12

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Re: m/n where n=0
« Reply #4 on: August 15, 2011, 02:23:22 am »

Computers, on the other hand, may have exceptions when dividing by zero. Again, the reason is that you can't really compute with undefined numbers. Some cases will return infinity (positive or negative) or a special value (NaN) but each of these works pretty badly in calculations if they weren't expected.

It's important to note here that these are just industry standard methods for dealing with division by zero, and do not necessarily reflect any mathematical fact.

Yep.
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Vector

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Re: m/n where n=0
« Reply #5 on: August 15, 2011, 02:40:39 am »

So, I just got off a high from winning an argument about dividing by zero, and I'm kind of wondering if there is any particular reason why you can't divide by zero. As far as I know, it is simply undefined, but no particular law of mathematics states that you can't. If anyone who knows more about math than me (not hard) knows of a law that specifies that you cannot, please tell me what the law is, for future reference.

Here's a hint: mathematical laws are made up based on what gives us the most interesting systems to work in.

The system (ring) where you can divide by zero is extremely boring.  Here are its components: 0.

Therefore, you cannot divide by zero.
« Last Edit: August 15, 2011, 03:03:21 am by Vector »
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Max White

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Re: m/n where n=0
« Reply #6 on: August 15, 2011, 02:59:56 am »

*Mumble mumble* Approaching a line *mumble* limits *gruff profanity*
Well, that doesn't really count, but it is the best you can get. Your sort of dividing by everything but zero.

Grek

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Re: m/n where n=0
« Reply #7 on: August 15, 2011, 03:13:36 am »

14/2 = 14*2^-1 = 7 because 7*2 = 14

14/0 = 14*0^-1 = ??? because X*0 =/= 14 for all values of X.

Thus, 0^-1 is undefined.
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Vector

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Re: m/n where n=0
« Reply #8 on: August 15, 2011, 03:32:21 am »

I would simplify it down to the statements that

a. 0 is defined s.t. X*0 = 0*X = 0 for all X in ring R (axioms of ring)
b. m/n = p when np = m
c. X/0 = p, X =/= 0 has 0p = X; contradiction, the construction does not exist for n = 0, given the existence of a unique unity.

This way is most beautiful to me, though you may feel free to disagree =)
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Grek

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Re: m/n where n=0
« Reply #9 on: August 15, 2011, 03:34:37 am »

That definition is unelegant, as it requires that you know what the terms "ring" "axiom" and "unity" mean.
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Simmura McCrea

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Re: m/n where n=0
« Reply #10 on: August 15, 2011, 03:36:10 am »

Let me give you a more practical example. You have, say, 10 of your largest denomination coins, whatever they are. Dollars, probably. To divide by 2, you'd arrange them into two piles, giving 5 in each. 10/2=5. Now try to divide them into 0 piles.
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Vector

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Re: m/n where n=0
« Reply #11 on: August 15, 2011, 03:42:21 am »

That definition is unelegant, as it requires that you know what the terms "ring" "axiom" and "unity" mean.

And your method is inelegant because it uses numbers.

Bad ones ::)  Why not use 1 instead of 7?

... Sorry.  I am not doing happy nonsense today.
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Max White

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Re: m/n where n=0
« Reply #12 on: August 15, 2011, 03:44:39 am »

Let me give you a more practical example. You have, say, 10 of your largest denomination coins, whatever they are. Dollars, probably. To divide by 2, you'd arrange them into two piles, giving 5 in each. 10/2=5. Now try to divide them into 0 piles.
*Purchase cookie for ten coins*
*Places all remaining coins in zero piles*

So the answer must be cookie!

Simmura McCrea

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Re: m/n where n=0
« Reply #13 on: August 15, 2011, 03:46:31 am »

Let me give you a more practical example. You have, say, 10 of your largest denomination coins, whatever they are. Dollars, probably. To divide by 2, you'd arrange them into two piles, giving 5 in each. 10/2=5. Now try to divide them into 0 piles.
*Purchase cookie for ten coins*
*Places all remaining coins in zero piles*

So the answer must be cookie!
Hella expensive cookie, bro. Could buy a lot of cookies for a tenner over here, but you'd still end up with change.
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