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Jul - General Chat - A new revolution in math! New poll - New thread - New reply
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Lyskar
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Posted on 01-31-11 11:24:24 PM Link | Quote
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Metal_Man88's Post
? / 0 is actually plus or minus infinity if you use limits, but that of course means 'undefined' as it doesn't really work for anything sensible.

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Drag
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Posted on 01-31-11 11:30:16 PM Link | Quote
1958
Originally posted by Grey
-J wouldn't be the smallest number, since making a number negative doesn't reduce its size. It just means it's of the greatest scalar magnitude in the negative direction.

The smallest number is zero.

...wow, ok, I just got served.

but anyway, x/0 is undefined, but when you think about what division is (how many times does X go into Y?), 0 goes into any number an infinite amount of times, so the result would be positive or negative infinity (depending on signs). The only thing is that there's really no reason to have anything divided by 0, so it's just undefined.

However, I can tell you that 0 goes into X a maximum of J times.

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Posted on 01-31-11 11:36:59 PM Link | Quote
And this is why infinity isn't considered a number. If it's treated the same as other numbers, you start getting stuff like:

∞ + 1 = ∞
-∞     -∞
1 = 0

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devin

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Posted on 01-31-11 11:44:46 PM (last edited by devin at 01-31-11 08:47 PM) Link | Quote
Originally posted by Drag
However, I can tell you that 0 goes into X a maximum of J times.

How? You can't divide two integers and get an even larger number.

edit: oh, "maximum of". dooooh

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Sine
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Posted on 01-31-11 11:47:10 PM (last edited by Sine at 01-31-11 08:48 PM) Link | Quote
It's an enigma
----------------------------------------------------
On a side note, let's just look at...

1+1+1+1+1+1... +1+1+1+1 <-- Let's say that there are J of these. But we're not going to call it J. Let's call it...

(1+1+1+1+1+...) + (... +1+1+1+1), where you dont know how many 1's are contained in either of these. Certainly they're both less than J, so they can be themselves.

Now let's add 1

(1+1+1+1+1+...) + (... +1+1+1+1) + 1.
(1+1+1+1+1+...) + (... +1+1+1+1+1)

I just made a number bigger than J

But J is supposed to be the biggest number! D:

Oh god Drag, what the hell did I just do?

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FieryIce

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Posted on 02-01-11 12:18:11 AM Link | Quote
Originally posted by Sine
On a side note, let's just look at...

1+1+1+1+1+1... +1+1+1+1 <-- Let's say that there are J of these. But we're not going to call it J. Let's call it...

(1+1+1+1+1+...) + (... +1+1+1+1), where you dont know how many 1's are contained in either of these. Certainly they're both less than J, so they can be themselves.

Now let's add 1

(1+1+1+1+1+...) + (... +1+1+1+1) + 1.
(1+1+1+1+1+...) + (... +1+1+1+1+1)

I just made a number bigger than J

But J is supposed to be the biggest number! D:

Oh god Drag, what the hell did I just do?


But, if you solve the parentheses first, you end up with: J + 1 = J

Am I right?

Sine
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Posted on 02-01-11 12:20:45 AM Link | Quote
It's an enigma
----------------------------------------------------
Originally posted by FieryIce
But, if you solve the parentheses first, you end up with: J + 1 = J

Am I right?
Not at all! We arent solving it that way. We're not even solving it.

We're just stating it as a number. Kinda like 2/1 = 4/2 and etc.

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FieryIce

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Posted on 02-01-11 12:25:21 AM Link | Quote
Originally posted by Sine
Originally posted by FieryIce
But, if you solve the parentheses first, you end up with: J + 1 = J

Am I right?
Not at all! We arent solving it that way. We're not even solving it.

We're just stating it as a number. Kinda like 2/1 = 4/2 and etc.


Ahh, I missed your point! You did make a number bigger than J. But, could it be that J is not a constant, but work as a variable kind of like "Universe" works in Boolean Logic? Sorry, I find this very interesting but I'm not good at all in math. ._.
Drag
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Posted on 02-01-11 12:25:58 AM Link | Quote
1959
Originally posted by Sine
On a side note, let's just look at...

1+1+1+1+1+1... +1+1+1+1 <-- Let's say that there are J of these. But we're not going to call it J. Let's call it...

(1+1+1+1+1+...) + (... +1+1+1+1), where you dont know how many 1's are contained in either of these. Certainly they're both less than J, so they can be themselves.

Now let's add 1

(1+1+1+1+1+...) + (... +1+1+1+1) + 1.
(1+1+1+1+1+...) + (... +1+1+1+1+1)

I just made a number bigger than J

But J is supposed to be the biggest number! D:

Oh god Drag, what the hell did I just do?

You just made J again.

What you're doing is the summation of (1) from 1 to (J - x), and then you're adding the summation of (1) from (J - x) to J to it, which makes J. Then you're adding 1 to it, which means you're doing J + 1, and that's defined as J + 1 = J.

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Sine
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Posted on 02-01-11 12:27:15 AM (last edited by Sine at 01-31-11 09:29 PM) Link | Quote
It's an enigma
----------------------------------------------------
Originally posted by Drag
You just made J again.

What you're doing is the summation of (1) from 1 to (J - x), and then you're adding the summation of (1) from (J - x) to J to it, which makes J. Then you're adding 1 to it, which means you're doing J + 1, and that's defined as J + 1 = J.
No I didnt.

I took two numbers who are both less than J, but whose sum is greater than that of J

Never did I use J anywhere in the equation

In fact

I can take any variation on (1+1+1...) < J and (1+1+...) < J, and produce you two numbers whose sum is greater than that of J

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Drag
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Posted on 02-01-11 12:34:57 AM Link | Quote
1960
Originally posted by Sine
I took two numbers who are both less than J, but whose sum is greater than that of J

Never did I use J anywhere in the equation

Sorry, I misunderstood.

None the less, it's impossible to come up with a number that's greater than J, so what you've actually done is:

X + Y + 1 < J

You mentioned that X and Y were both < J, but you have a fallacy in your equation, because there is no number greater than J. Thus, you added X + Y together, but you still ended up with a number less than J, and then you added 1 to it, making another number that is still less than J.

The fact that J wasn't in the equation means that you cannot come up with a number greater than or equal to J.

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Sine
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Posted on 02-01-11 12:39:53 AM (last edited by Sine at 01-31-11 09:43 PM) Link | Quote
It's an enigma
----------------------------------------------------
Originally posted by Drag
Sorry, I misunderstood.

None the less, it's impossible to come up with a number that's greater than J, so what you've actually done is:

X + Y + 1 < J

You mentioned that X and Y were both < J, but you have a fallacy in your equation, because there is no number greater than J. Thus, you added X + Y together, but you still ended up with a number less than J, and then you added 1 to it, making another number that is still less than J.

The fact that J wasn't in the equation means that you cannot come up with a number greater than or equal to J.
Why is it impossible to come up with a number greater than J, I already did.

You're essentially saying that in order for J to exist, that X and Y are dependent on each other, IE that Y < J - X - 1, which violates the fact that I picked X and Y

And what happens when I want a+b+c+d+...n, for any number (or in this case, J number of variables) where a...n are all greater than 1 (in fact, make them all variations of 2 with random decimal points).

You're running yourself into a barrel of contradictions good sir

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FieryIce

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Posted on 02-01-11 12:40:53 AM Link | Quote 


J x 1 = J

J x 2 = J



J x 1 = J x 2



J/J x 1 = J/J x 2



1 = 2
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Posted on 02-01-11 12:46:29 AM (last edited by FirePhoenix at 01-31-11 09:47 PM) Link | Quote
Fuck this math i have a test tomorrow in not-math

But the test has math on it

WHICH IS IT? A MATH TEST OR A NOT-MATH TEST?

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Drag
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Posted on 02-01-11 12:58:13 AM Link | Quote
1961
Sine: I'm not really contradicting myself at this point, since I've had ample time to think about the J theory.
The first part of the theorem is that there is no number greater than J. It's absolutely impossible to increase the value of J. That's why you cannot come up with a number bigger than J.

If you have J number of variables, and those variables are all 1 or greater, then you still end up with J, because you explicitly declared that there are "J" number of variables. Since we know that there can be no number bigger than J, we know that the answer is going to be J, since you're just doing J + Constant again.

At the same time, if you have J variables, and those variables are all less than 1, then the result will always be less than J.

If you have J variables that you added together, and then you add some arbitrary constant to it, then you're just doing J + n again, and what's that defined as?

FieryIce: Not quite, J * 1 = J, and J * 2 = J. You have to evaluate J before you do any other operations.

J * 1 = J * 2
1J = 2J
(1J)/J = (2J)/J
(J)/J = (J)/J
1 = 1

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Posted on 02-01-11 01:00:21 AM Link | Quote
Originally posted by Metal_Man88
? / 0 is actually plus or minus infinity if you use limits, but that of course means 'undefined' as it doesn't really work for anything sensible.
Actually, you can make it be anything you want with a combination of simple algebra and limits. For example, say I have this equation:

f(x) = (x² + x - 2) / (x - 1)

In this case, the equation becomes division by zero when x = 1. The limit of f(x) as x approaches 1 is 3.

Using math like that, I can define division by zero to be whatever I feel like it should be.

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Posted on 02-01-11 01:01:39 AM Link | Quote

Originally posted by Joe
Originally posted by Metal_Man88
? / 0 is actually plus or minus infinity if you use limits, but that of course means 'undefined' as it doesn't really work for anything sensible.
Actually, you can make it be anything you want with a combination of simple algebra and limits. For example, say I have this equation:

f(x) = (x² + x - 2) / (x - 1)

In this case, the equation becomes division by zero when x = 1. The limit of f(x) as x approaches 1 is 3.

Using math like that, I can define division by zero to be whatever I feel like it should be.

In that case you're dividing 0 by 0 (sort of), which is even more of a special case than dividing random non-zero numbers by zero.

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Posted on 02-01-11 01:07:51 AM Link | Quote
Call me crazy, but I'm pretty sure J+1 is sqrt(2)<45°, not J.

Electrical engineering

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Lyskar
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Posted on 02-01-11 01:51:12 AM Link | Quote
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Metal_Man88's Post
Originally posted by Imajin
Originally posted by Joe
Originally posted by Metal_Man88
? / 0 is actually plus or minus infinity if you use limits, but that of course means 'undefined' as it doesn't really work for anything sensible.
Actually, you can make it be anything you want with a combination of simple algebra and limits. For example, say I have this equation:

f(x) = (x² + x - 2) / (x - 1)

In this case, the equation becomes division by zero when x = 1. The limit of f(x) as x approaches 1 is 3.

Using math like that, I can define division by zero to be whatever I feel like it should be.

In that case you're dividing 0 by 0 (sort of), which is even more of a special case than dividing random non-zero numbers by zero.


Dividing 0 by 0 requires L'Hospital's rule, if it's even applicable. Sometimes it isn't, making a mess.


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Posted on 02-01-11 02:01:52 AM Link | Quote
0/0 is indeterminate, which is still undefined but not the same type of undefined as, say, 1/0. Something like 1/0 will always go to ∞ or -∞ when you take its limit, whereas 0/0's limit could be a number.

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