Why can not you divide any number by zero?

Question

i was thinking how strange the number zero is! if you multiply something by it, you get zero. but why can not you divide any thing by zero? why?

Answer 1

i.e.

5/0=?
?x0=5

Algebraic interpretation

It is generally regarded among mathematicians that a natural way to interpret division by zero is to first define division in terms of other arithmetic operations. Under the standard rules for arithmetic on integers, rational numbers, real numbers and complex numbers, division by zero is undefined. Division by zero must be left undefined in any mathematical system that obeys the axioms of a field. The reason is that division is defined to be the inverse operation of multiplication. This means that the value of \textstyle\frac{a}{b} is the solution x of the equation bx = a whenever such a value exists and is unique. Otherwise the value is left undefined.

For b = 0, the equation bx = a can be rewritten as 0x = a or simply 0 = a. Thus, in this case, the equation bx = a has no solution if a is not equal to 0, and has any x as a solution if a equals 0. In either case, there is no unique value, so \textstyle\frac{a}{b} is undefined. Conversely, in a field, the expression \textstyle\frac{a}{b} is always defined if b is not equal to zero.

[edit] Fallacies based on division by zero

It is possible to disguise a special case of division by zero in an algebraic argument, leading to spurious proofs that 2 = 1 such as the following:

With the following assumptions:

0\times 1 = 0
0\times 2 = 0

The following must be true:

0\times 1 = 0\times 2

Dividing by zero gives:

\textstyle \frac{0}{0}\times 1 = \frac{0}{0}\times 2

Simplified, yields:

1 = 2\,

The fallacy is the implicit assumption that dividing by 0 is a legitimate operation with 0 / 0 = 1.

Although most people would probably recognize the above "proof" as fallacious, the same argument can be presented in a way that makes it harder to spot the error. For example, if 1 is denoted by x, 0 can be hidden behind x − x and 2 behind x + x. The above mentioned proof can then be displayed as follows:

(x-x)x = x^2-x^2 = 0\,
(x-x)(x+x) = x^2-x^2 = 0\,

hence:

(x-x)x = (x-x)(x+x)\,

Dividing by x-x\, gives:

x = x+x\,

and dividing by x\, gives:

1 = 2\,

The "proof" above requires the use of the distributive law. However, this requirement introduces an asymmetry between the two operations in that multiplication distributes over addition, but not the other way around. Thus, the multiplicative identity element, 1, has an additive inverse, -1, but the additive identity element, 0, does not have a multiplicative inverse.

Answer 2

zero is a weird number. in fact, it never even used to exist. some people just had to have some number to show the non-existence of objects. It is that none-existent nature of zero that gives it this nature.

The way I look at dividing by zero is the way the people who came up with zero would have: trying to split up property.

Just think about it: say you have one acre of land to split up. You can divide it into 1, 1000 parts, or even 10000 parts. But when you take that field and try to divide it zero times, it isn't possible. there is still the land. The minimum ways that field can be divided is into 1.

More or less, it is the same with the numbers, though when using pure numbers you can use decimals, infinitely close to zero, or even negative. zero is just an interesting concept. It is just nothing.

The rules in math for ordinary math would be that multiplication and division are reciprical functions, that is they cancel one another out. for example, 4*x= 5, then x = 5/4. when it comes to zero though, this cannot apply. Whereas 5*0 = 0, 0/0 doesn't equal 5.

Answer 3

The truth is, you have to distinguish between two cases:

1: Any number except for 0 divided by 0:
This will have no correct solution at all: not infinity, not anything, because there is *no* number that you can multiply by 0 to get any number other than 0 itself. 23/0 is asking for a number that does not exist.

2: 0 divided by 0:
The problem here is the opposite: Every solution is correct. *Every* number can be multiplied by 0 to get 0. So if you wanted, you could say that 0 divided by 0 is the set of all numbers, and that would be true. In practical terms, though, that is not an answer that gets you anywhere worthwhile. It's the same as if I had answered your question by saying. "The answer to your question is the answer to your question". Technically true, but utterly useless in terms of telling you something that you didn't know.

Answer 4

Think about multiplication. 3 x 5 = 15. What that really is, is adding either (5+5+5) or (3+3+3+3+3).
Now consider 0 x 5 --> (0+0+0+0+0) or (5 added together 0 times)

Divsion: 15 / 5 = 3. There is no simple way to write this like above, but 5 goes into 15, 3 times.
Now 15 / 0 --> Basically 0 can't go 'into' any number, hence undefined. Look at it this way. $15 divided by 5 people is $3 for each person. What would $15 be divided by 0 people?

Answer 5

"i be responsive to that individuals declare that any extensive style (besides 0) divided through 0 equals to infinity yet once you multiply infinity and nil, you will get 0" those human beings are incorrect. In prevalent arithmetic (field arithmetic of the particular extensive style line), branch through 0 is undefined. continually. You _can_ divide through 0 interior the arithmetic of the particular projective line, and that's actual the end results of dividing any extensive style through 0 is infinity. So in that context, yeah, a million/0 = ?. even regardless of the indisputable fact that: in that comparable context, the manufactured from ? and nil is undefined. "So my question is, shouldn't a extensive style (besides 0) divides through 0 substitute into quotient 0 and the rest is the extensive style ( eg. a million/0=0 the rest a million )." No. look at how integer quotients are defined. a / b = c the rest d the place c is the _largest_ integer such that bc ? a. Now, you want to declare that if a=a million and b=0 then c=0 and d=a million, yet ask your self: are you able to discover a c extra advantageous than 0 such that bc ? a? nicely, how approximately c=a million? bc = 0*a million = 0 ? a How approximately c=2? bc = 0*2 = 0 ?a How approximately c=3? bc = 0*3 = 0 ?a you spot the place this is going. there is not any best integer c such that bc ? a, considering the fact that that inequality holds for _all_ integers.

Answer 6

Multiplication is repeated addtion where as Division is repeated subtraction.

Let us take simple division. 6/2 means,
6-2=4, 4-2=2, 2-2=0 and you did 3 finite steps of repeated subtraction which is the quotient 3 and reminder is 0.

Let us take 6/0. You will go in an uncountably infinite steps 6-0=6, 6-0=6,... etc. and I don't see any end to this. Do you? That is why you can not divide by 0, and if you divide you will get infiity.

Pleas let me know if you have any more questions.

Thanks,
Arivoli

Answer 7

OK - I'll be the first to admit I'm not a math genius so I don't understand any of the explanations I've seen here. To me it's simpler.

When you devide for example 12 by the number 2 you are dividing 12 into two equal parts. What happens when you try to divide it by 0? Can you divide something into 0 parts? No.

Can it get any easier?

Answer 8

8 / 4 = 2 because 4 x 2 = 8
therefore:
5 / 0 = some number (x)
therefore:
0 times x = 5

but no value for x multiplied by 0 would ever give you five, b/c like you said any number multiplied by zero gives you zero

Answer 9

Suppose we can divide a non-zero number m by zero. It means there is a number which is a quotient. Let it be n
=> m/0 = n
=> m = 0 x n = 0
=> m = 0
But m is non-zer. Thus there is a contradiction.
Hence, division by zero is not permissible.

Answer 10

Because dividing zero by any number gives you an infinity