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- Thread starter l46kok
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Because this is what may happen in the real world

http://photos-974.ll.facebook.com/photos-ll-sf2p/v143/24/40/711375974/n711375974_204141_7854.jpg [Broken]

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- #3

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l46kok, think on how do you

Now imagine b is zero. What number, multiplied by zero, gives you some particular number a?

If a is not zero too, then there is not such a number. No number multiplied by zero gives non-zero, that's why a / 0 has no result if a is non-zero.

And if a is also zero, then

Hope this helps.

- #4

arildno

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Sure it is:

1. Having no reference to division at all, we may prove by standard properties of real numbers that for any real "a", a*0=0.

2. If therefore, for real numbers b and for real numbers c, the division b:c shall equal a number k so that c*k=b, it follows that for any non-zero b, we cannot in general let c=0,

since we already have proven that..c*k=0.

Therefore, in order to maintain consistency of our mathematical operations we see that division by 0 is prohibited

- #5

Hurkyl

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Because division is defined to operate on nonzero denominators -- your question is like asking why we use the word "blue" to decribe light with a wavelength of 465 nm.

There are lots of reasons why this definition is a useful one -- the other posts demonstrate some bad things that we would face if we attempted to define a new division operator that allowed division by zero.

- #6

Zurtex

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Because division is defined to operate on nonzero denominators -- your question is like asking why we use the word "blue" to decribe light with a wavelength of 465 nm.

There are lots of reasons why this definition is a useful one -- the other posts demonstrate some bad things that we would face if we attempted to define a new division operator that allowed division by zero.

To be fair that's not entirely true. While you can look it that way, just dryly, saying that "because that's the way it's defined".

A more interesting way to look at it is to consider the Field of real numbers, which have a certain set of properties that must obey. Then it's non-trivial to show that 0 has no multiplicative inverse, you tackle it much the way arildno has.

- #7

arildno

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I think Hurkyl was very fair, Zurtex; because he is right.

Although we shouldn't throw out the idea of mathematical consistency, I'm sure we could throw out the distributive law of multiplication, say, and preserve a "division" concept applicable for zero.

(But then, multiplication is no longer what it once was..)

Essentially, therefore, the ultimate justification for why division by 0 is prohibited is DEFINING the operation in such a way that it remains consistent/meaningful with whatever other axioms are held to be true.

If we choose to change those axioms themselves, we might not need to exclude 0 from being a divisor in our definition of division.

Although we shouldn't throw out the idea of mathematical consistency, I'm sure we could throw out the distributive law of multiplication, say, and preserve a "division" concept applicable for zero.

(But then, multiplication is no longer what it once was..)

Essentially, therefore, the ultimate justification for why division by 0 is prohibited is DEFINING the operation in such a way that it remains consistent/meaningful with whatever other axioms are held to be true.

If we choose to change those axioms themselves, we might not need to exclude 0 from being a divisor in our definition of division.

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- #8

Zurtex

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My quible, if any, was to do the the analogy:

'your question is like asking why we use the word "blue" to decribe light with a wavelength of 465 nm.'

The reason it's called blue is arbitrary, we could call it green, we could it elephant, we could call it awoejhropwiheroiuwheifuwheoifugwoeifugilqhwb4ihfbghfwiherbfwihebfiwhegbf. The reason our definition of division doesn't allow for for divide by 0 isn't arbitrary, 0 has special properties in a field and it wouldn't make sense for us to choose some other qualitatively different number.

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Because that’s how fields are defined.

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Zurtex

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Because that’s how fields are defined.

I know you

http://mathworld.wolfram.com/FieldAxioms.html

Where devision is defined through the multiplicative inverse, the Field Axioms then

It can also nicely be shown that a field becomes inconsistent if you allow 0 to have a multiplicative inverse.

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I think my answer is the best one to this question because it lets him go as deep as he wants it has the: “Because that’s how we define it” element but if he wants to know WHY we define it that way he need to study fields.I know youcandefine a field like that, but I was always taught to define a field by these:

http://mathworld.wolfram.com/FieldAxioms.html

Where devision is defined through the multiplicative inverse, the Field Axioms thenimplydivision by 0 isn't possible because 0 has no multiplicative inverse. Sorry I may seem a little over the top about this, I just don't feel in this case it's a very sufficient answer for someone who is trying to learn something new.

It can also nicely be shown that a field becomes inconsistent if you allow 0 to have a multiplicative inverse.

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CRGreathouse

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0 + 1 = 1 + 0 = 1 (additive identity)

Thus 1 + 1 = 0 because otherwise there is no additive inverse.

Distributing, (1 + 1) * 0 = (1 * 0) + (1 * 0).

1 * 0 = 0 (multiplicative identity)

So (1 + 1) * 0 = 0 + 0 = 0 (additive identity for the last step)

Of course we also know that 1 + 1 = 0, so (1 + 1) * 0 = 0 * 0.

If 0 has a multiplicative inverse, it must be 0 (1 doesn't work by multiplicative identity). Thus 0 * 0 = 1.

So 0 = (1 + 1) * 0 = 1. But we know that 1 and 0 are distinct.

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i dunno what the big deal is. when we say:

[tex] Q = \frac{N}{D} [/tex]

we are saying that quotient

do you accept first that zero times anything is zero? it has to be if the distributive property is true. so if you accept that

[tex] 0 \times Q = 0 [/tex]

then there is no way for

[tex] D \times Q = N [/tex]

if

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CRGreathouse

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[tex]\{a\}\oplus\{b\}=a\oplus b[/tex] where [itex]\oplus[/itex] is +, -, or *.

Then [itex]\{a\}/\{0\}=\mathbb{R}[/itex] if [itex]a=0[/itex] and [itex]\{a\}/\{0\}=\emptyset[/itex] otherwise.

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Let's try another way.

Then [itex]\{a\}/\{0\}=\mathbb{R}[/itex] if [itex]a=0[/itex] ..

this is potentially true (but i always worry about what it is in the limit, but there doesn't appear to be a limit here), but there is no way we can tell what

- #16

morphism

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?this is potentially true (but i always worry about what it is in the limit, but there doesn't appear to be a limit here), but there is no way we can tell whatparticularreal number [itex]\{a\}/\{0\}[/itex] is. it could be anything, making [itex]\{a\}/\{0\}[/itex] pretty meaningless whetherais zero or not.

What CRGreathouse did was define operations on sets. What he wrote down isn't "potentially true", it's absolutely true. Under his definitions, {a}/{0} is not a real number, it's a set.

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If you partition 16 into equal parts of size 2, then you would need 8 such equal partitions. However, what happens when you try to form partitions of size 0. How many would you need? An infinite number of them (and intuitively, even an infinite number of them may not be enough, which leads to indeterminate forms -- but that's for another rainy day). Unfortunately, infinity isn't a number -- it's a concept (at least as far as I am concerned; though it's considered a number in the extended reals).

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CRGreathouse

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What CRGreathouse did was define operations on sets. What he wrote down isn't "potentially true", it's absolutely true. Under his definitions, {a}/{0} is not a real number, it's a set.

Yeah, my operations are just direct and indirect images in disguise. I hoped the form of it would show that even in that extension there can be no single element that will work.

You could try adding in ur-elements, but I think you'd find the same problems -- it you have division by zero your other properties fall apart. You could have x / 0 = 3 or x / 0 = [itex]\mathcal{U}\!\mathcal{R}_1[/itex] (some arbitrary ur-element), but then you have to abandon the idea that 0x = 0 for all x, which means you can't have additive inverses. By fixing up a small hole in multiplicative inverses you blow a huge one in the additive inverses -- is it worth it?

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If you partition 16 into equal parts of size 2, then you would need 8 such equal partitions. However, what happens when you try to form partitions of size 0. How many would you need? An infinite number of them (and intuitively, even an infinite number of them may not be enough, which leads to indeterminate forms -- but that's for another rainy day). Unfortunately, infinity isn't a number -- it's a concept (at least as far as I am concerned; though it's considered a number in the extended reals).

This is not a mathematical argument, and it assumes too much. It may be easier to think about it in terms of metaphors, but that is not mathematics, it is philosophy!

Why use "laymen's terms" when we have an exact construction and contradiction in proof by CRGreathouse?

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This is not a mathematical argument, and it assumes too much. It may be easier to think about it in terms of metaphors, but that is not mathematics, it is philosophy!

I do not claim that this is a mathematical argument; hence "in layman's terms." Metaphors? Partitions are well known notions in mathematics (e.g. combinatorics). Given a/b, where a and b are positive real numbers, division is equivalent to partitioning a into equal parts of size b, and then counting the number of parts. There is nothing metaphorical about this approach. Philosophy? To someone lacking the mathematical background (esp. someone without the mathematical background needed to understand that the [tex]\mathbb{R}[/tex] is a field), the other approaches naturally raise the question: why are fields important?

Why use "laymen's terms" when we have an exact construction and contradiction in proof by CRGreathouse?

My point is that one can deduce inconsistencies with lesser tools if division by zero is allowed. I have nothing against rigorous proofs, and as a mathematician, encourage rigorous arguments. However, proofs are generally the end result of mathematical understanding. (How often is it that you solve a problem by jumping straight to the proof?) I prefer to use "layman's terms" because the OP's question can be addressed without resorting to more abstract objects such as fields; basic intuition about partitions and positive real numbers is sufficient for showing that inconsistencies occur if we extend the notion of division to include 0. After that comes a proof.

- #21

CRGreathouse

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CRGreathouse said:Let's try another way.

Then [itex]\{a\}/\{0\}=\mathbb{R}[/itex] if [itex]a=0[/itex]..

this is potentially true (but i always worry about what it is in the limit, but there doesn't appear to be a limit here), but there is no way we can tell whatparticularreal number [itex]\{a\}/\{0\}[/itex] is. it could be anything, making [itex]\{a\}/\{0\}[/itex] pretty meaningless whetherais zero or not.

It took me a few readings, but I think I see what you're getting at. I'm not claiming that {a}/{0} is a real number ([itex]\{a\}/\{0\}\in\mathbb{R}[/itex]) but that {a}/{0} is the set of real numbers ([itex]\{a\}/\{0\}=\mathbb{R}[/itex]).

In fact, I showed that there's no real number to which {a}/{0} corresponds:

[tex]\not\exists e\in\mathbb{R}:\{a\}/\{0\}=\{e\}[/tex]

- #22

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?

What CRGreathouse did was define operations on sets. What he wrote down isn't "potentially true", it's absolutely true. Under his definitions, {a}/{0} is not a real number, it's a set.

It took me a few readings, but I think I see what you're getting at. I'm not claiming that {a}/{0} is a real number ([itex]\{a\}/\{0\}\in\mathbb{R}[/itex]) but that {a}/{0} is the set of real numbers ([itex]\{a\}/\{0\}=\mathbb{R}[/itex]).

In fact, I showed that there's no real number to which {a}/{0} corresponds:

[tex]\not\exists e\in\mathbb{R}:\{a\}/\{0\}=\{e\}[/tex]

yup, i needed to be prodded to read it again, carefully. what CRG said, with the notation he used (which i was not being careful with) is precisely true.

[tex] \{a\}/\{0\}=\mathbb{R} [/tex] if [tex] a =0 [/tex]

hell, it could be [tex]\{a\}/\{0\}=\mathbb{C}[/tex] if [tex] a = 0 [/tex]

which means that [itex]\{a\}/\{0\}[/itex] is sorta meaningless since it cannot resolve to any smaller set of values other than

[tex] \{a\}/\{0\}=\emptyset [/tex] if [tex] a \ne 0 [/tex]

which says that [itex]\{a\}/\{0\}[/itex] can't be anything.

One case, [itex] a = 0 [/itex], says that [itex]\{a\}/\{0\}[/itex]

Let

[tex]a = b[/tex]

Multiply through by

[tex]a^2 = ab[/tex]

Subtract [tex]b^2[/tex]

[tex]a^2 - b^2 = ab - b^2[/tex]

Factor both sides

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

Divide out [itex](a - b)[/itex]

[tex]a + b = b[/tex]

Observing that [itex]a = b[/itex]

[tex]b + b = b[/tex]

Combine like terms on the left

[tex]2b = b[/tex]

Divide by

[tex]2 = 1 \,[/tex]

and if 2 = 1, then 1 = 0, then [itex] \infty [/itex] = 0, "yes" = "no", "right" = "wrong", Hitler was right all along, and George W. Bush really

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Because division is defined to operate on nonzero denominators .

Good point. With regard to fractions, it's imposible to have a zero deniminator. You can not have any value "out of" zero.

- #24

Hurkyl

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Just to introduce more perspective...

In commutative algebra, there is a procedure called "localization". Given a ring*R* (such as the integers, rationals, reals, complexes, or other interesting things), you can select a set of elements *S* that's closed under multiplication, and then you can compute a new ring [itex]S^{-1}R[/itex] allows you to divide by any of the elements in *S*.

For example, if we let*R* be the integers and *S* be the positive integers, then [itex]S^{-1} R[/itex] is isomorphic to the rational numbers.

In this construction, you**are** to put zero in *S*. This has a rather drastic effect on arithmetic, however; it makes all elements of [itex]S^{-1} R[/itex] equal.

Let me say that again, with an example: if we start with the integers, and we construct a new ring from the integers where division by zero is allowed, then in this new ring, all numbers are equal. For example, [itex]0 = 1[/itex] in this ring.

In commutative algebra, there is a procedure called "localization". Given a ring

For example, if we let

In this construction, you

Let me say that again, with an example: if we start with the integers, and we construct a new ring from the integers where division by zero is allowed, then in this new ring, all numbers are equal. For example, [itex]0 = 1[/itex] in this ring.

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