Angel_Light

03-14-2007, 12:01 AM

Is it possible to do the distance formula Graal? Does graal have square root symbol? It has powers "^2".

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Angel_Light

03-14-2007, 12:01 AM

Is it possible to do the distance formula Graal? Does graal have square root symbol? It has powers "^2".

Chandler

03-14-2007, 12:05 AM

temp.objectDistance = ((this.x ^ 2) + (this.y ^ 2)) ^ 0.5;

:)

:)

Angel_Light

03-14-2007, 12:07 AM

Thank you very much! =D

DustyPorViva

03-14-2007, 12:07 AM

Square root is ^.5

Anyways, to further on what Chandler said, this.dx and this.dy should be this.dx and this.dy. Though it really doesn't matter as the variables are what you make of them, but for understanding purposes, you're getting the distance from delta's.

Anyways, to further on what Chandler said, this.dx and this.dy should be this.dx and this.dy. Though it really doesn't matter as the variables are what you make of them, but for understanding purposes, you're getting the distance from delta's.

Gambet

03-14-2007, 12:14 AM

The root of a number is the same thing as saying that same number to the power of the exponent it's being multipled to inside the root over the degree of the root.

How that works?

Well, when you rationalize a root, if the root has no outside degree showing, then it's automatically a square root (degree of 2). If you a number inside that square root that does not have an exponent (i.e. doesn't have a power), then the power is automatically 1, since any number to the power of 1 is that same number.

So, by knowing that, you can apply what I said earlier to realize that, for example, the square root of 2 is the same thing as saying 2 to the power of 1/2, or 0.5 (decimal form).

To write the power of a number, you use this symbol: ^

So, it would be written as:

2^0.5 or 2^1/2

How that works?

Well, when you rationalize a root, if the root has no outside degree showing, then it's automatically a square root (degree of 2). If you a number inside that square root that does not have an exponent (i.e. doesn't have a power), then the power is automatically 1, since any number to the power of 1 is that same number.

So, by knowing that, you can apply what I said earlier to realize that, for example, the square root of 2 is the same thing as saying 2 to the power of 1/2, or 0.5 (decimal form).

To write the power of a number, you use this symbol: ^

So, it would be written as:

2^0.5 or 2^1/2

Angel_Light

03-14-2007, 01:04 AM

a=1

b=1

a=b

aČ=ab

aČ+aČ=aČ+ab

2aČ=aČ+ab

2aČ-2ab=aČ+ab-2ab

2aČ-2ab=aČ-ab

2(aČ-ab)=1(aČ-ab)

1=2

yay

also

Does 0.9 = 1?

1/3 +1/3 +1/3 =1

Put it into decimal, note the 3 in 0.3 is repeating.

0.3 + 0.3 + 0.3 = 0.9

So does 0.9 = 1.

b=1

a=b

aČ=ab

aČ+aČ=aČ+ab

2aČ=aČ+ab

2aČ-2ab=aČ+ab-2ab

2aČ-2ab=aČ-ab

2(aČ-ab)=1(aČ-ab)

1=2

yay

also

Does 0.9 = 1?

1/3 +1/3 +1/3 =1

Put it into decimal, note the 3 in 0.3 is repeating.

0.3 + 0.3 + 0.3 = 0.9

So does 0.9 = 1.

JkWhoSaysNi

03-14-2007, 01:09 AM

Yes.

0.999.... repeating = 1.

0.999.... repeating = 1.

Chris

03-14-2007, 01:31 AM

Yes.

0.999.... repeating = 1.

LOL

0.999.... repeating = 1.

LOL

JkWhoSaysNi

03-14-2007, 01:35 AM

what?

Was Angel_Light joking? It's hard to infer humour from text :(

Was Angel_Light joking? It's hard to infer humour from text :(

DustyPorViva

03-14-2007, 01:43 AM

1+1=11

Sum41Freeeeek

03-14-2007, 02:16 AM

derr

Tolnaftate2004

03-14-2007, 04:26 AM

2(aČ-ab)=1(aČ-ab)

1=2

aČ-ab = 0, so removing them is division by 0, which is a math foul.

1=2

aČ-ab = 0, so removing them is division by 0, which is a math foul.

Rapidwolve

03-14-2007, 04:32 AM

Is it possible.

Anything is possible

Anything is possible

Gambet

03-14-2007, 04:44 AM

Anything is possible

Hence my custom user title (:

Hence my custom user title (:

Kristi

03-14-2007, 01:18 PM

aČ-ab = 0, so removing them is division by 0, which is a math foul.

well, if they were divided out, it would be infinity=infinity, so it is all good.

well, if they were divided out, it would be infinity=infinity, so it is all good.

napo_p2p

03-14-2007, 07:47 PM

well, if they were divided out, it would be infinity=infinity, so it is all good.

Then, it's not 1=2 anymore ;).

Then, it's not 1=2 anymore ;).

Tolnaftate2004

03-14-2007, 11:29 PM

well, if they were divided out, it would be infinity=infinity, so it is all good.

0/0 does not definitely equal infinity.

In fact, to make Taylor/Maclaurin series work, 0/0 is finite (likely 1).

On another note, if you're going to use ^1/2, you may want to think again (PEMDAS). ^(1/2)...

0/0 does not definitely equal infinity.

In fact, to make Taylor/Maclaurin series work, 0/0 is finite (likely 1).

On another note, if you're going to use ^1/2, you may want to think again (PEMDAS). ^(1/2)...

Kristi

03-15-2007, 01:33 AM

0/0 does not definitely equal infinity.

In fact, to make Taylor/Maclaurin series work, 0/0 is finite (likely 1).

On another note, if you're going to use ^1/2, you may want to think again (PEMDAS). ^(1/2)...

Umm, 0 was never divided by 0 in that equation.

In fact, to make Taylor/Maclaurin series work, 0/0 is finite (likely 1).

On another note, if you're going to use ^1/2, you may want to think again (PEMDAS). ^(1/2)...

Umm, 0 was never divided by 0 in that equation.

Tolnaftate2004

03-15-2007, 05:56 AM

Umm, 0 was never divided by 0 in that equation.

2(aČ-ab)=1(aČ-ab)

2(aČ-ab)/(aČ-ab)=1(aČ-ab)/(aČ-ab)

recall that (aČ-ab) = 0

gives us 2*0/0 = 1*0/0

Hey, what do you know.

2(aČ-ab)=1(aČ-ab)

2(aČ-ab)/(aČ-ab)=1(aČ-ab)/(aČ-ab)

recall that (aČ-ab) = 0

gives us 2*0/0 = 1*0/0

Hey, what do you know.

Kristi

03-15-2007, 10:16 AM

2(aČ-ab)=1(aČ-ab)

2(aČ-ab)/(aČ-ab)=1(aČ-ab)/(aČ-ab)

recall that (aČ-ab) = 0

gives us 2*0/0 = 1*0/0

Hey, what do you know.

Well either way, by order of operations in that case, the 2 and 1 would have both been multiplied by 0 before being divided by zero. So 0/0 = 0/0, which I do believe in modern math is concidered 0.

2(aČ-ab)/(aČ-ab)=1(aČ-ab)/(aČ-ab)

recall that (aČ-ab) = 0

gives us 2*0/0 = 1*0/0

Hey, what do you know.

Well either way, by order of operations in that case, the 2 and 1 would have both been multiplied by 0 before being divided by zero. So 0/0 = 0/0, which I do believe in modern math is concidered 0.

Angel_Light

03-15-2007, 01:55 PM

I wonder what would happen IF we put those equation in GS2 and allow the language to solve it .=o

DustyPorViva

03-15-2007, 01:59 PM

"OK, now what you gotta do is go down the road past the old Johnson place. You're gonna find two roads, one parallel and one perpendicular. Now keep going until you come to a highway that bisects it at a 45-degree angle. Solve for x."

Twinny

03-15-2007, 02:27 PM

"OK, now what you gotta do is go down the road past the old Johnson place. You're gonna find two roads, one parallel and one perpendicular. Now keep going until you come to a highway that bisects it at a 45-degree angle. Solve for x."

*Curls into fetal position and sucks thumb*

*Curls into fetal position and sucks thumb*

Gambet

03-15-2007, 09:07 PM

"OK, now what you gotta do is go down the road past the old Johnson place. You're gonna find two roads, one parallel and one perpendicular. Now keep going until you come to a highway that bisects it at a 45-degree angle. Solve for x."

How can one road be parallel to itself? :rolleyes:

How can one road be parallel to itself? :rolleyes:

Angel_Light

03-15-2007, 09:56 PM

How can one road be parallel to itself? :rolleyes:

Kinda the point XD lol

Kinda the point XD lol

Skyld

03-15-2007, 10:45 PM

How can one road be parallel to itself? :rolleyes:

Uncertainty theory? >_<

Uncertainty theory? >_<

Tolnaftate2004

03-16-2007, 12:10 AM

Well either way, by order of operations in that case, the 2 and 1 would have both been multiplied by 0 before being divided by zero. So 0/0 = 0/0, which I do believe in modern math is concidered 0.

I can rearrange the numbers to do division first... If they're both multiplied by 0, I'd say the statement reads 0=0 and leave it at that. There is no infinity in either case. On top of that, 0/0 could be anything in C. Also, the taylor series for e (and sin and cos probably, along with others, I really haven't thought about it) needs 0/0 to be 1 to work. Oh, and TI-89s replace 0/0 with 1... ??? :confused:

I can rearrange the numbers to do division first... If they're both multiplied by 0, I'd say the statement reads 0=0 and leave it at that. There is no infinity in either case. On top of that, 0/0 could be anything in C. Also, the taylor series for e (and sin and cos probably, along with others, I really haven't thought about it) needs 0/0 to be 1 to work. Oh, and TI-89s replace 0/0 with 1... ??? :confused:

Gambet

03-16-2007, 12:23 AM

You can't divide any number by 0, you can only divide 0 by a number.

What are you guys talking about? :confused:

What are you guys talking about? :confused:

Tolnaftate2004

03-16-2007, 12:30 AM

You can't divide any number by 0, you can only divide 0 by a number.

This sort of thing is dealt with all the time in calculus. But I think that is a discussion for somewhere else.

This sort of thing is dealt with all the time in calculus. But I think that is a discussion for somewhere else.

Kristi

03-16-2007, 03:30 AM

This sort of thing is dealt with all the time in calculus. But I think that is a discussion for somewhere else.

Yes, Gambet, stop thinking so algebraically. It all has to do with approaching limits.

for f(x) = 1/x

.5 = 1/2

1 = 1/1

2 = 1/.5

10 = 1/.1

....

as x approaches 0, f(x) gets infinitely larger and larger, thus equating to infinity.

in the case of f(x) = 0/x

0 = 0/2

0 = 0/1

0 = 0/.5

0 = 0/.1

....

as x approaches 0, f(x) stays 0, so 0/0 = 0.

Yes, Gambet, stop thinking so algebraically. It all has to do with approaching limits.

for f(x) = 1/x

.5 = 1/2

1 = 1/1

2 = 1/.5

10 = 1/.1

....

as x approaches 0, f(x) gets infinitely larger and larger, thus equating to infinity.

in the case of f(x) = 0/x

0 = 0/2

0 = 0/1

0 = 0/.5

0 = 0/.1

....

as x approaches 0, f(x) stays 0, so 0/0 = 0.

Gambet

03-16-2007, 03:45 AM

Yeah I'm taking Algebra II Honors this year, all I can do is think algebraically -_-

Angel_Light

03-16-2007, 02:05 PM

Yes, Gambet, stop thinking so algebraically. It all has to do with approaching limits.

for f(x) = 1/x

.5 = 1/2

1 = 1/1

2 = 1/.5

10 = 1/.1

....

as x approaches 0, f(x) gets infinitely larger and larger, thus equating to infinity.

in the case of f(x) = 0/x

0 = 0/2

0 = 0/1

0 = 0/.5

0 = 0/.1

....

as x approaches 0, f(x) stays 0, so 0/0 = 0.

*Joins Twinny in the fetal position while sucking my thumb*

for f(x) = 1/x

.5 = 1/2

1 = 1/1

2 = 1/.5

10 = 1/.1

....

as x approaches 0, f(x) gets infinitely larger and larger, thus equating to infinity.

in the case of f(x) = 0/x

0 = 0/2

0 = 0/1

0 = 0/.5

0 = 0/.1

....

as x approaches 0, f(x) stays 0, so 0/0 = 0.

*Joins Twinny in the fetal position while sucking my thumb*

Twinny

03-16-2007, 02:44 PM

*Joins Twinny in the fetal position while sucking my thumb*

I left all those stuff at school. I never ever want to do another primitive function ever again. I did all right in calculus and trigonemtry and such but.....damn if I will ever use it. Or will ever be able to remember it >_<

I left all those stuff at school. I never ever want to do another primitive function ever again. I did all right in calculus and trigonemtry and such but.....damn if I will ever use it. Or will ever be able to remember it >_<

Kristi

03-17-2007, 12:03 AM

I can rearrange the numbers to do division first... If they're both multiplied by 0, I'd say the statement reads 0=0 and leave it at that. There is no infinity in either case. On top of that, 0/0 could be anything in C. Also, the taylor series for e (and sin and cos probably, along with others, I really haven't thought about it) needs 0/0 to be 1 to work. Oh, and TI-89s replace 0/0 with 1... ??? :confused:

The TI-89 is wrong. The only way you can describe 0/0 equating to one is because of the algebraic rule that a/a is always equal to one, but that implies that a is a nonzero number. If you think about what division means in algebra, it establishs a ratio. if you did 2/5, then you would have ,4:1, if you did 1/2, then you would have .5:1, if you did 5/2, then you would have 2.5:1, etc etc. but what if you have a/0? You could never establish it as b:1. a/a as a ratio is 1:1, unless it is 0, which is why a/a when a=0 does not equal 1. The TI-89 incorrectly uses an algebraic rule in a situation where it cannot apply.

The only way to solve 0/0 is to use a limit approach, which makes it equal to 0, and which is the right answer defined by modern mathematics .

The TI-89 is wrong. The only way you can describe 0/0 equating to one is because of the algebraic rule that a/a is always equal to one, but that implies that a is a nonzero number. If you think about what division means in algebra, it establishs a ratio. if you did 2/5, then you would have ,4:1, if you did 1/2, then you would have .5:1, if you did 5/2, then you would have 2.5:1, etc etc. but what if you have a/0? You could never establish it as b:1. a/a as a ratio is 1:1, unless it is 0, which is why a/a when a=0 does not equal 1. The TI-89 incorrectly uses an algebraic rule in a situation where it cannot apply.

The only way to solve 0/0 is to use a limit approach, which makes it equal to 0, and which is the right answer defined by modern mathematics .

napo_p2p

03-17-2007, 01:12 AM

Oh, and TI-89s replace 0/0 with 1... ??? :confused: Wierd. Mines says 'undef' for 0/0.

Tolnaftate2004

03-17-2007, 02:14 AM

Wierd. Mines says 'undef' for 0/0.

When you graph in 3D, I think it says "0/0 replaced with 1." I've only tinkered with an 89 for a little while, I don't actually own one, though.

The only way to solve 0/0 is to use a limit approach, which makes it equal to 0, and which is the right answer defined by modern mathematics .

I can make several limits to equate 0/0, and each of them gives a different result.

e: in fact,

lim (x,y) -> (0,0) y/x (does not exist) but evaluated individually gives C.

When you graph in 3D, I think it says "0/0 replaced with 1." I've only tinkered with an 89 for a little while, I don't actually own one, though.

The only way to solve 0/0 is to use a limit approach, which makes it equal to 0, and which is the right answer defined by modern mathematics .

I can make several limits to equate 0/0, and each of them gives a different result.

e: in fact,

lim (x,y) -> (0,0) y/x (does not exist) but evaluated individually gives C.

Kristi

03-17-2007, 03:58 AM

When you graph in 3D, I think it says "0/0 replaced with 1." I've only tinkered with an 89 for a little while, I don't actually own one, though.

I can make several limits to equate 0/0, and each of them gives a different result.

e: in fact,

lim (x,y) -> (0,0) y/x (does not exist) but evaluated individually gives C.

It doesn't matter what your calculator says, 0/0 is 0

I can make several limits to equate 0/0, and each of them gives a different result.

e: in fact,

lim (x,y) -> (0,0) y/x (does not exist) but evaluated individually gives C.

It doesn't matter what your calculator says, 0/0 is 0

Angel_Light

03-17-2007, 06:53 AM

I use a TI 83+ so I know it's not as powerful as a TI 89 but mine gives me undefined and when I used my friends TI 92 it said 0/0 is one... so.... I just lost myself in thought >_<

Torankusu

03-17-2007, 08:47 AM

Zero is simple. If you do not have anything, then you cannot receive anything.

For example, If you split nothing (0) between (/) two (2) people, then they both have two nothings, which, equals 0.

The same is going to hold true if you divide nothing between no one. It will not even occur and there will still be nothing. You will not create something and end up with one (1) object.

For you math wizzes, I'm sure you know zero literally and mathematically means "nothing."

Oddly enough though, if you do 0^0 you will end up with one.

For example, If you split nothing (0) between (/) two (2) people, then they both have two nothings, which, equals 0.

The same is going to hold true if you divide nothing between no one. It will not even occur and there will still be nothing. You will not create something and end up with one (1) object.

For you math wizzes, I'm sure you know zero literally and mathematically means "nothing."

Oddly enough though, if you do 0^0 you will end up with one.

Kristi

03-17-2007, 12:59 PM

Zero is simple. If you do not have anything, then you cannot receive anything.

For example, If you split nothing (0) between (/) two (2) people, then they both have two nothings, which, equals 0.

The same is going to hold true if you divide nothing between no one. It will not even occur and there will still be nothing. You will not create something and end up with one (1) object.

For you math wizzes, I'm sure you know zero literally and mathematically means "nothing."

Oddly enough though, if you do 0^0 you will end up with one.

once again its a calculator error. 0^0 is equal to 0.

f(x) = 0^x

f(2) = 0^2 = 0

f(1) = 0^1 = 0

f(.5) = 0^.5 = 0

f(.1) = 0^.1 = 0

f(.01) = 0^.01 = 0

....

as you can see, as you approach 0, the value stays 0, so 0^0 = 0. you cannot apply the algebraic rule that a^0 = 1 unless its a nonzero number.

For example, If you split nothing (0) between (/) two (2) people, then they both have two nothings, which, equals 0.

The same is going to hold true if you divide nothing between no one. It will not even occur and there will still be nothing. You will not create something and end up with one (1) object.

For you math wizzes, I'm sure you know zero literally and mathematically means "nothing."

Oddly enough though, if you do 0^0 you will end up with one.

once again its a calculator error. 0^0 is equal to 0.

f(x) = 0^x

f(2) = 0^2 = 0

f(1) = 0^1 = 0

f(.5) = 0^.5 = 0

f(.1) = 0^.1 = 0

f(.01) = 0^.01 = 0

....

as you can see, as you approach 0, the value stays 0, so 0^0 = 0. you cannot apply the algebraic rule that a^0 = 1 unless its a nonzero number.

Tolnaftate2004

03-17-2007, 06:37 PM

It doesn't matter what your calculator says, 0/0 is 0

I'm not using a calculator.

So,

lim x->0 of 0/x = 0

lim x->0+ of x/0 = infinity

lim x->0- of x/0 = -infinity

lim x->0 of x/x = 1

lim x->0 of (x/2)/x = 1/2

lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:

lim x-> 0 of 0^x = 0

lim x-> 0 of x^0 = 1

All "modern algebra" does is lead me to believe that 0/0 is some set that includes {0,1} at least.

I'm not using a calculator.

So,

lim x->0 of 0/x = 0

lim x->0+ of x/0 = infinity

lim x->0- of x/0 = -infinity

lim x->0 of x/x = 1

lim x->0 of (x/2)/x = 1/2

lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:

lim x-> 0 of 0^x = 0

lim x-> 0 of x^0 = 1

All "modern algebra" does is lead me to believe that 0/0 is some set that includes {0,1} at least.

Angel_Light

03-17-2007, 10:29 PM

My brain hurts now. >_<

Kristi

03-18-2007, 12:06 AM

I'm not using a calculator.

So,

lim x->0 of 0/x = 0

lim x->0+ of x/0 = infinity

lim x->0- of x/0 = -infinity

lim x->0 of x/x = 1

lim x->0 of (x/2)/x = 1/2

lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:

lim x-> 0 of 0^x = 0

lim x-> 0 of x^0 = 1

You are using limits to describe finite concepts. We already know 0 can be divided into 0 parts. Limit definitions are solid when finite definitions do not exist, eg diving by zero. Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

So,

lim x->0 of 0/x = 0

lim x->0+ of x/0 = infinity

lim x->0- of x/0 = -infinity

lim x->0 of x/x = 1

lim x->0 of (x/2)/x = 1/2

lim x->0 of nx/x = n (n!=0)

I know how to show 0 = 0, but how about the rest of these? How is infinity = 0? If n != 0, how is it 0, then? 0^0 or 0/0 cannot simply be described with one number.

An additional note:

lim x-> 0 of 0^x = 0

lim x-> 0 of x^0 = 1

You are using limits to describe finite concepts. We already know 0 can be divided into 0 parts. Limit definitions are solid when finite definitions do not exist, eg diving by zero. Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

Tolnaftate2004

03-18-2007, 12:23 AM

You are using limits to describe finite concepts.

A limit can be applied anywhere. A limit only exists when the limit is finite, anyway. This statement really confuses me.

Limit definitions are solid when finite definitions do not exist, eg diving by zero.

But you keep saying 0/0 = 0. 0 is neither -infinity nor infinity, so it is finite.

Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

Each of the limits above are valid for evaluating 0/0. The "modifying operation" is rather ambiguous in this case. 0/0 could be split up to 0*1/0, or any other number of ways and we'd have a few operations to choose from.

Depending on how we approach 0/0, we can get infinitely different answers. Consider the equation (3x^2+nx)/x. At x=0, we have 0/0, and there is a removable discontinuity, but we can still evaluate the limit and it gives us n.

The function happens to be continuous. If we say that 0/0 = 0, we've broken the continuity.

A limit can be applied anywhere. A limit only exists when the limit is finite, anyway. This statement really confuses me.

Limit definitions are solid when finite definitions do not exist, eg diving by zero.

But you keep saying 0/0 = 0. 0 is neither -infinity nor infinity, so it is finite.

Generally speaking, you apply the limit to the modifying operation, eg division, raising to a power, etc.

Each of the limits above are valid for evaluating 0/0. The "modifying operation" is rather ambiguous in this case. 0/0 could be split up to 0*1/0, or any other number of ways and we'd have a few operations to choose from.

Depending on how we approach 0/0, we can get infinitely different answers. Consider the equation (3x^2+nx)/x. At x=0, we have 0/0, and there is a removable discontinuity, but we can still evaluate the limit and it gives us n.

The function happens to be continuous. If we say that 0/0 = 0, we've broken the continuity.

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