Dividing by zero

[houserichichi]

Again, I hate to be a nitpicker, but

because 0^0 is undefined

isn't true...we say that 0^0 is indeterminate which is very different. Minor detail, but can throw you way off the higher up the math chain you go.

Locked by Rob: because a duplicate topic exists.

[jinydu]

Ok, so what is the difference between undefined and indeterminate? And why do we say that 0^0 is indeterminate but not undefined?

[houserichichi]

Well this is the easy answer and if you'd prefer I write more I can do later, but it's getting late here so I'll give you the quicky answer :wink: I'm not sure how deep into math you've gone yet, so I'll keep it general for now.

The general rule of thumb goes by saying something is undefined if it is not allowable by the "rules" of arithmetic. However, we say something is indeterminate if there is no way of telling what the actual value would be.

It sounds a little vague, but think of it this way

undefined -> no possible answer because it's illegal to do
indeterminate -> we don't know what the answer is

[jinydu]

Well this is the easy answer and if you'd prefer I write more I can do later, but it's getting late here so I'll give you the quicky answer :wink: I'm not sure how deep into math you've gone yet, so I'll keep it general for now.

The general rule of thumb goes by saying something is undefined if it is not allowable by the "rules" of arithmetic. However, we say something is indeterminate if there is no way of telling what the actual value would be.

It sounds a little vague, but think of it this way

undefined -> no possible answer because it's illegal to do
indeterminate -> we don't know what the answer is

In that case, it seems that 0^0 should be considered undefined, because:

0^0 = 0^(1-1)
0^0 = (0^1)/(0^1)
0^0 = 0/0

This should be undefined, because it requires dividing by zero, which is not allowed (the most we can do is take the limit as the derivative approaches 0). Of course, its also also indeterminate, otherwise the derivative of any function at any x-value would always be the same number.

[houserichichi]

Ok now I'm really going to bed after this one...BUT both 0^0 and 0/0 are indeterminate. Think of taking limits lim(a,b)->(0,0) for 0^b and a^0...depending on which direction we come in from results in different answers. Same goes for a/0 and 0/b. It's simple multivariable calc, but it's a very deep result. Many folks prefer to set 0^0 = 1 for convenience, but it's technically wrong.

[Keiji]

a/0 = infinity
0/b = zero
0/0 = all real numbers

'Nuff said (tm).

[jinydu]

Getting back to tetration, here are the first few terms of the Taylor series for x^x about x = 1:

x^x ~= 1 + (x-1) + (x-1)^2 + (1/2)(x-1)^3 + (1/3)(x-1)^4

Expanding and simplifying:

x^x ~= (1/3)x^4 - (5/6)x^3 + (3/2)x^2 - (5/6)x + (5/6)

For x = 1.1, this approximation gives x^x =~ 1.110533333333333333..., quite close to the actual (provably irrational) value of 1.11053424...

[RQ]

a/0 = infinity
0/b = zero
0/0 = all real numbers

'Nuff said (tm).

a/0=infinity? where do you get that from?

0/0=all real numbers? The simplest proof agains that is if 0/0=1 then

[0/0]2=[1]2

0/0=2

1=2?

[houserichichi]

it's off topic from tetration so I'll keep it brief...

a / 0 is undefined for any a != 0. It diverges to infinity as a -> 0.

0 / 0 can take on a range of values, it just depends what direction we come from to get to it. It's an indeterminate form.

And in your proof that 1 = 2, you can't cancel a 2 on the left and a 1 on the right. :wink:

On to tetration.........(didn't mean for it to go this off topic, sorry!)

[RQ]

What am I canceling?

The initial assumption was that 0/0=1, multiply both sides by 2, and replace 0/0 with 1. What canceling?

Edit: I wasn't proving that 1=2, I was disproving that 0/0=all numbers x.

[Keiji]

[0/0]2=[1]2 <-- Multiplied both sides by 2

0/0=2 <-- Divided left side by 2, and right side by 1

You can't do that.

[jinydu]

This is rapidly turning into a digression about dividing by 0 and 0^0. Bobxp, please split this into another thread or merge it into the 0/0 thread.

[Keiji]

I've split it, but I can't merge it into the other thread so I'll lock that one and quote the stuff here:

1/0=0/0

How many of you believe me? Nobody? Ok take a look at this:

0/0=0/0

1-1/0=0/0

[1/0]-[1/0]=0/0

[1/0][2/2]-[1/0]=0/0

[2/0]-[1/0]=0/0

1/0=0/0 QED

When you start dealing with division by zero you get nonsense answers...it's still an indeterminate form which is why things don't look right. When you multiply (1/0) by (2/2) you're just using the natural laws or arithmetic that we were taught in grade school to perform the operation...however, 2/0 has no meaning (nor 1/0) so the proof doesn't show anything...it shows a set of symbols are equal, not numbers.

(1) The simplest proof agains that is if 0/0=1 then

(2) [0/0]2=[1]2

(3) 0/0=2

(4) 1=2?

I figured it belonged here instead...I thought you were multiplying the left side by 2 and the right side by 2 on line (2)...because if that's the case your next line (3) requires that you divide by 2 on the left and divide by 1 on the right.

0/0 can be ANY number, depending on the situation...you just showed that 1=2 by assuming that 0/0=1 which is not a valid assumption. Had you assumed that it equal something else, you could have had a different result, so your proof shows nothing more than arithmetic done under a false assumption.

Remember that dividing by zero allows you to prove statements that are obviously false. This is a classic:

Assume a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)*(a-b) = b*(a-b)
a+b = b
a+a = a
2a = a
2 = 1

[Keiji]

1/0=0/0

How many of you believe me? Nobody? Ok take a look at this:

0/0=0/0

1-1/0=0/0

[1/0]-[1/0]=0/0

[1/0][2/2]-[1/0]=0/0

[2/0]-[1/0]=0/0

1/0=0/0 QED

2/0 = infinity
1/0 = infinity
Therefore 2/0 = 1/0
Therefore 2/0 - 1/0 = 0
Therefore, using your algebra, 0=0/0

Hmm. I seem to have just pwn3d myself. Lol.

[houserichichi]

well now that it's got its own topic... :wink: I don't feel so bad talking about it as much!

1/0 does not equal infinity. As we approach from the right it tends to positive infinity and tends to negative infinity as we approach from the left. So, from a geometric standpoint there's no definition of 1/0 at all...algebraically we'd be looking for the solution (value of x) of

0x = 1

so we require a value of x when multiplied by 0 that gives 1. Since, I assume, we're dealing with the reals (which is a field, for all those who know what that means), we must have 0x = 0 for all x, there is no solution to the equation above. However, from the simple statement above (0x = 0) we see that there are infinitely many solutions, that is, x can take any real numbered value.

[jinydu]

Getting back to tetration, here are the first few terms of the Taylor series for x^x about x = 1:

x^x ~= 1 + (x-1) + (x-1)^2 + (1/2)(x-1)^3 + (1/3)(x-1)^4

Expanding and simplifying:

x^x ~= (1/3)x^4 - (5/6)x^3 + (3/2)x^2 - (5/6)x + (5/6)

For x = 1.1, this approximation gives x^x =~ 1.110533333333333333..., quite close to the actual (provably irrational) value of 1.11053424...

Bobxp, could you please merge this post back into the tetration thread?

[Keiji]

ARGH!

I CANNOT MERGE POSTS! I ALREADY SAID THAT!

[RQ]

[0/0]2=[1]2 <-- Multiplied both sides by 2

0/0=2 <-- Divided left side by 2, and right side by 1

You can't do that.

How is it divided left side by 2 and right side by 1?

Initial statement:

0/0=1

Step One: Multiply BOTH sides by 2:

0/0=2

Step Two: Replace 0/0 with 1:

1=2

If this is correct, my grandma is a chimney sweeper.

[RQ]

1/0=0/0

How many of you believe me? Nobody? Ok take a look at this:

0/0=0/0

1-1/0=0/0

[1/0]-[1/0]=0/0

[1/0][2/2]-[1/0]=0/0

[2/0]-[1/0]=0/0

1/0=0/0 QED

2/0 = infinity
1/0 = infinity
Therefore 2/0 = 1/0
Therefore 2/0 - 1/0 = 0
Therefore, using your algebra, 0=0/0

Hmm. I seem to have just pwn3d myself. Lol.

[2/0]-[1/0]=0 Where do you get that statement from?
2/0=infinity? Any proof?

[RQ]

So, from a geometric standpoint there's no definition of 1/0 at all...algebraically we'd be looking for the solution (value of x) of

0x = 1

so we require a value of x when multiplied by 0 that gives 1. Since, I assume, we're dealing with the reals (which is a field, for all those who know what that means), we must have 0x = 0 for all x, there is no solution to the equation above. However, from the simple statement above (0x = 0) we see that there are infinitely many solutions, that is, x can take any real numbered value.

0*x=0 and x=0/0 are not the same equations, unless you are assuming that 0/0=1 which disproves the outcome.

[RQ]

Remember that dividing by zero allows you to prove statements that are obviously false. This is a classic:

Assume a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)*(a-b) = b*(a-b)
a+b = b
a+a = a
2a = a
2 = 1

That's only if you evaluate 0/0 as 1.

[houserichichi]

That's only if you evaluate 0/0 as 1.

Not exactly - he's got to multiply both sides on the right by the inverse of (a-b) = 0 to cancel it...that's where the error occurs because in the reals on cannot divide by zero.

a^2 - b^2 = ab - b^2
(a+b)*(a-b) = b*(a-b)

Just so you know Jinydu, I think I remember you saying you were taking a math degree...get used to writing things in the proper order...the second line should read

(a-b)*(a+b) = (a-b)*b

in which case I was wrong above - you have to multiply on the left by the inverse of (a-b), not the right...but hey, that's something you'll get used to doing the higher up you go. (If someone uses the excuse that we're not using the reals, then this gets rid of that problem...works for any field now). God I feel ocd about details...I'll cut it out now :wink:

[Keiji]

2/0=infinity? Any proof?

See the circular number line theory.

[RQ]

Not exactly - he's got to multiply both sides on the right by the inverse of (a-b) = 0 to cancel it...that's where the error occurs because in the reals on cannot divide by zero.

0 has no inverse, and if you are talking about a reciprocal, then again, x/0 would mean that 0*x/0 makes 0/0=1 which is as you said a contradiction. The equations are here:

Step 1)Statement:
a=b

Step 2)Multiply each side by a
a^2=ab

Step 3)Subtract each side by b^2
a^2-b^2=ab-b^2

Step 4)Factor
(a+b)(a-b)=b(a-b)

Step 5)Divide each side by (a-b)
a+b=b

Step 6)Substitute a for b or vice versa
b+b=b or a+a=a

Step 7)Simplify
2b=b

Step Evaluate
2=1

Now here's the downfall of the whole argument. In step 5 we have the equation: (a+b)(a-b)=b(a-b)

We divide that by (a-b) to get

[(a+b)(a-b)]/(a-b)=[b(a-b)]/(a-b) which equals

[a+b]*[(a-b)/(a-b)]=b*[(a-b)]/(a-b)

Since a=b, we can rewrite this as

(a+b)*[0/0]=b*[0/0]

If saying that the above equation and a+b=b are the same then this means that 0/0 is evaluated as 1, which is shown to be false. If however 0/0 is evaluated as 0, then we have no problems.

[RQ]

2/0=infinity? Any proof?

See the circular number line theory.

The only circular number line theory I know is the number line curled up in a circle showing that negative infinity equals positive infinity, but it neither says why or how the number line is curved in a circle.

[jinydu]

That's only if you evaluate 0/0 as 1.

Not exactly - he's got to multiply both sides on the right by the inverse of (a-b) = 0 to cancel it...that's where the error occurs because in the reals on cannot divide by zero.

a^2 - b^2 = ab - b^2
(a+b)*(a-b) = b*(a-b)

Just so you know Jinydu, I think I remember you saying you were taking a math degree...get used to writing things in the proper order...the second line should read

(a-b)*(a+b) = (a-b)*b

in which case I was wrong above - you have to multiply on the left by the inverse of (a-b), not the right...but hey, that's something you'll get used to doing the higher up you go. (If someone uses the excuse that we're not using the reals, then this gets rid of that problem...works for any field now). God I feel ocd about details...I'll cut it out now :wink:

But shouldn't it be the same? Multiplication is commutative.

[houserichichi]

But shouldn't it be the same? Multiplication is commutative.

In the reals, yeah it is...but I figured just in case someone said "well what if we weren't in the reals" and threw a silly example where the multiplication isn't commutative...it's rare that anyone knows to do that, but it's good habit to get used to proving them wrong in all scenarios (assuming you prefer algebra over analysis and geometry). When I was younger that was my favourite false proof in the world - I stumped many of my friends with it...and I recall getting a teacher for a while too. Oh the joys of being a dork.

[RQ]

I don't think that imaginary numbers multiplied from the left side make the outcome different than being multiplied from the right side, but if you're so sure that they do, why don't you give us an example?

Since people read math (in the US at least) from left to right, it's more comfortable to put the b on the left side than on the right.

[houserichichi]

I don't think that imaginary numbers multiplied from the left side make the outcome different than being multiplied from the right side

No, the complex numbers are a field as well, and in fields multiplication is always commutative. If, however, we used matrices instead, it's normally the case that AB does not equal BA...they don't form a field, but again you can't "divide" by zero (the zero matrix, that is). So if you want to solve for A in AB = 0, for instance, you have to multiply the RIGHT by the inverse of B whereas for BA = 0 you'd have to multiply on the LEFT by the inverse of B. That's the simplest example I could think of off the top of my head. (Let's assume we were using square matrices too - to make the discussion easier.)[/quote]

[PWrong]

I knew matrices were noncommutative, but I didn't think complex numbers would be. I read somewhere once that there was a type of complex number with four, um, dimensions, like a + bi + cj + dk, called quaternions or something. It said that quaternions were also non-commutative.

I'm doing a maths degree next year as well. Now I'm interested in the difference between algebra and analysis. Do people often prefer one over the other?