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Report this Nov. 24 2002, 4:11 pm

Why is a number divided by zero not defined?


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Report this Nov. 24 2002, 4:31 pm

Well that’s a great question I think.

To keep it simple just think about polynomials in algebra.

Look at a degree 0 or a constant polynomial. Like the function of x = 0. When we look at that we have y = 0 so graph that. That line is horizontal and in this case is right on the x axis. It really has no m or slope because it is a constant (that is why we call it a constant polynomial), but lets still try to figure out its slope. Pick two points like (1,0) & (0,0)

So we have
m = (delta y) / (delta x)
m = (0-0)/(1-0) = 0/1 = 0
So we want to break up 0 into 1 area. Of course we have zero so we just want to break that evenly into 1. That seems logical to allways get 0.

When we graph a degree 0 polynomial it can only be horizontal, but what about vertical? We cannot have ’y = some #’ to have it completely vertical [up & down] it’s not possible. This in it self has to get an light bulb lighting for us. The only way to do this is ’x = some #’, but that does not really fit in a more general look at a polynomial. Lets say x = 0 in this case it would actually be the same line as the y axis. Now lets try to find the slope. Lets look at (0,2) & (0,0)

So we have
m = (2-0)/(0-0) = 2/0 = undefined! (?)
You might be asking in the first place how can a line like this not have sloped it’s not flat like flat on the ground. I know where you are getting at when you look at this. But besides that think about the # 2 split evenly into 0 areas. How is that possible to split a # into zero

I recommend just playing around just a bit with constant polynomials and finding out the slope and looking at the graph or line that they make on the coordinate plane. If you first see that a vertical line does not fit ’y = _’ which is a big thing, finding out the slope, and just thinking about dividing a # by zero. Then I think you can make a lot of connections to why it is undefined and should be.

I hope that helped a bit or helped as a starter.

Master Q


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Report this Nov. 24 2002, 5:36 pm

Just came back thought I would add something to what I said.

One thing about dividing a # by 0 is to think about the fact that 4/2 = 2 and 2*2 = 4 so x/0 = nothing times 1 would get you right back to 0. For more of a algebraic way to say it a/b = x so b*x = a. Meaning it would not work.
Thats probably the best way to look at it, but by looking at constnat polynomials it also would get you on the right track.

Master Q


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Report this Nov. 25 2002, 7:32 pm

Or to put it more simply, try dividing something into zero groups... voila! Impossible.


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Report this Nov. 25 2002, 7:39 pm

Undefined can mean many things:

It can mean "need more info" like infinity - infinity. It is not always zero, it is undefined because we don’t know what infinity were talking about. For example all odd numbers (1,3,5,7,9...) is infinity, right? And all even and odd numbers (1,2,3,4,5...) is also infinity. But when we take our second infinity and subtract our first infinity (1,2,3,4,5...) - (1,3,5,7,9...) we get all even numbers (2,4,6,8,10...) and that is infinity! So in this case infinity - infinity = infinity! That is why it is "undefined". This is also true for 0 / 0 it isn’t always 1. (different story)

It can also mean "we’re not going to tell you in this level of math" or "you will learn the real answer later". An example is the square root of a negative #. In algebra I they said it is undefined but in Algebra II now it is actually imaginary numbers! (different story)

Or it can mean "we’re not really sure" or "you will never need to know anyway so don’t worry". This is the case with 1 / 0...

To mirror what Master_Q said, no # we know of when muliplied by 0 equals 1! Even infinity because all the odd #’s (1,3,5,7,9...) each by them self when multiplied by 0 equals 0 so the whole thing equals 0! So this new # which is equal to 1 / 0 has to be greater than infinity! . . .

Note: This shouldn’t make a lot of sense right away, I mean greater than infinity?!?! Just think about it. While it makes no cognitive sense it is logical.

. . . So we’ll make a new number and call it (for simplicity purposes) #. So 1 / 0 = #. And 2 / 0 = 2#. Therefore, # x 0 = 1! And 2# x 0 = 2! This # is outside the set of real numbers and impossible to comprehend visually. This also expliains how a vertical line’s slope doesn’t seem to work - it just doesn’t - there is no graphing #’s beyond visual comprehension! Really weird but the truth.

Note: If this just made no sense whatsoever then don’t be discouraged. I’ve been thinking about this for years so it takes a while to grasp the concepts.

Hope that helped!

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