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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

People often ask in maths class over the years why any number devided by 0 is no solution. They often ask why it isn't equal to 0, infinity or an imaginary number. Well I have a very simple answer to this.

If you divide any number like so:

6/3 = 2

You can find which number the numerator was by multiplying the denominator by the answer.

So y/3 = 2, y = 6.

y can only possibly be 6. There is no other possible solution.

y/3 = 0, y = 0.

Once again there is only one possible solution.

But in the case of of the following.

3/0 = x

The product of x and 0 does not give the numerator, 3. It gives 0. This does not follow our rule. So in the algabraic expression:

x/0 = y. Supposing y was magically given, x could could equal an infinite number of things. x/0 in this case would have more than one possible answer and thus, the problem yields no solution.

*"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."*

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Great!

And Simple!

**X'(y-Xβ)=0**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Here is another reason, albeit a slightly more advanced one.

When dealing with real numbers, we can talk about them as a field. One requirement to be a field is that there has to be multiplicative inverses (division), except for the additive identity (0). In other words, every real number has a multiplicative inverse besides 0.

So when you say x / 0, what you are really saying is, "x times the multiplicative inverse of 0". But 0 doesn't have a multiplicative inverse! So the statement makes no sense at all!

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

The multiplicative inverse of a number I assume you mean 1/number or reciprocal?

So if 3/0 is unclear, then why is 1/0 so clear?

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

What do you mean by unclear and clear?

In a field, the additive identity can't have a multiplicative inverse. 0 is the additive identity of the field of real numbers, so saying it has a multiplicative inverse "/0" makes the reals not a field.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Sorry, I don't have enough knowledge to understand where you are coming from.

I probably havn't read much on this subject, however, are you saying that 3/0's undefinability can be determined by 1/0's undefinability?

**igloo** **myrtilles** **fourmis**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,659

Multiplicative inverse of x: what you multiply by x to get 1. (=1/x or x-¹)

Additive inverse of x: what you add to x to get 0. (=-x)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Not exactly John. What I'm basically saying is that regardless of the number, the "/0" part just doesn't make sense.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

For anyone who is interested:

Prove that the additive identity can not have a multiplicative inverse in a field.

Proof: This proof is by contradiction. Let F be a field, and assume that the addititive identity, 0, has a multiplicative inverse. That is, 0a = a0 = 1, for some a ∈ F. For F to be a field, it must hold that F is a ring as well. Thus, the distributive property holds (as it must in all rings):

a(b + c) = ab + ac and (a + b)c = ac + bc

But consider the equation 0 + 0 = 0, which must hold because 0 is the additive identity. Applying some x ∈ F to both sides we get x(0 + 0) = x(0). And by the distributive property, x0 + x0 = x0. Since the field must be a group under addition, every element must have an additive inverse, and this includes x0. So x0 + x0 + (x0)-¹ = x0 + (x0)-¹, and so x0 + (0) = 0 which means x0 = 0.

The same argument can be used to show that 0x = 0. Thus, any element multiplied by 0 is 0. But field must also be an integral domain, which must have a non-zero multiplicative identity (called the unity). Thus, it must be that 0 ≠ 1. Contradiction.

∴ It must be that the additive identity does not have a multiplicative inverse. QED.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Ricky, when you said So x0 + x0 + (x0)-¹ = x0 + (x0)-¹, did you mean x0 + x0 - x0 = x0 - x0 ?

Also, why does field have to be integral domain?

Off the subject I was doing some reading on infinity (and as you can probably tell, I don't read a lot), I found it was very interesting how many different studies have been done, and rules that try to unify everything and make sense of it.

Over time, I am being slowly affected by you Ricky. You have me wondering if infinity really takes on no values and perhaps is disjoint from real numbers because as you approach infinity, as least linearly, whereever you are is always closer to zero than infinity, so it seems, unless you start believing in infinity/2 and that whole realm of varying sizes of infinity. Anyway, enough for now.

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

So x0 + x0 + (x0)-¹ = x0 + (x0)-¹, did you mean x0 + x0 - x0 = x0 - x0 ?

Yes, exactly! Subtraction is definited as adding the inverse of a number. Just like division is defined as multiplying the inverse of a number.

Over time, I am being slowly affected by you Ricky.

I guess that's a good thing. But don't take everything I say is fact. Remember, I make mistakes too, and trust me, I've been known to make some great ones.

so it seems, unless you start believing in infinity/2 and that whole realm of varying sizes of infinity

There are varying sizes of infinity, if you talk about it in the right context. For example, the set of natural numbers is infinitely large. But it is still smaller than the set of real numbers. And weirdly enough, even though the natural numbers are a proper subset of the integers, the integers and natural numbers are the same size.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

It seems Ricky's solution is a defined version with a same core...

**X'(y-Xβ)=0**

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

It also seems to me that 0/0 = anything and everything.

let a be some real number,

a * 0 = 0, a = 0/0

thus 0/0 = any and every real number. Not really a legitimate proof but its fun to pretend. :-)

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

a0 = 0 is true for any real number. But again, you are trying to divide by 0. In other words, you are multiplyinig by the multiplicative inverse of 0. Because the inverse of 0 doesn't exist, it just doesn't make sense. It's like asking what the color blue tastes like or naming a country south of the south pole.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Yeah I know, like I said its fun to pretend.

A logarithm is just a misspelled algorithm.

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