Question #fc10d

3 Answers
Dec 10, 2016

Answer:

I think this argument is up for debate, however personally believe it is actually #0#.

Explanation:

Clearly the two rules you have proposed contradict, however I would argue that no items shared among no people leaves nothing for anyone.

Dec 10, 2016

Answer:

Undefined/indeterminate

Explanation:

How to classify #0/0# depends both on context and intent.


First, as this question is in prealgebra, let's look at it without any reference to higher level mathematics. In this context, we should treat it as undefined.

When we perform the operation #x/y=x-:y#, it is exactly equivalent to multiplying #x# by the multiplicative inverse of #y^(-1)# of #y#, where #y^(-1)# is the value satisfying #yxxy^(-1) = 1#.

As there is no value which can be multiplied by #0# to produce #1#, #0# has no multiplicative inverse, and so multiplying by the multiplicative inverse of #0# is not possible. It is neither a finite value nor infinite, but rather is undefined, because there is no way to perform the operation.


As the question also mentions division by #0# resulting in infinity, let's also address how that can occur and what is meant when it does. To do so will require concepts from calculus, however as this is the prealgebra section, this answer will avoid using calculus terminology or assume prior calculus knowledge.

In calculus, we consider what happens when values get very large or very close to specific values. Rather than dividing by #0#, then, we can look at what happens when a dividend gets very close to #0#.

  • #1/1 = 1#
  • #1/(1/2) = 2#
  • #1/(1/10) = 10#
  • #1/(1/10000) = 10000#
    #...#

Notice that as the dividend gets closer to #0#, then quotient gets larger and larger. This pattern continues, and we can make the quotient larger than any given number by making the dividend close enough to #0#, i.e. the quotient becomes infinitely large as the dividend approaches #0#. Sometimes a calculus shorthand for this is writing #1/0 = oo#, even though we are not dividing by #0# nor is #oo# actually a number.

It gets trickier when we have #0# as a divisor, though.

  • #0/1 = 0#
  • #0/(1/2) = 0#
  • #0/(1/10) = 0#
    #...#

It seems if we just plug in #0# as the divisor, then the quotient is #0# every time, and so we should have #0/0=0#. But remember that we aren't actually dividing, but are looking at what happens very close to certain numbers. Consider another pattern:

  • #1/1 = 1#
  • #(1/2)/(1/2) = 1#
  • #(1/10)/(1/10) = 1#
    #...#

Both the divisor and dividend are getting close to #0#, but this time the quotient is #1# each time. Other patterns could lead to the quotient being any given finite number, or even growing infinitely large. In the case when we are considering the divisor and the dividend getting close to #0#, there isn't a general rule for how the quotient will behave, and so we say that #0/0# is an indeterminate form.


In general, different branches of math and physics treat #0/0# differently, depending on how they arrive there and the desired outcome.

Dec 14, 2016

Answer:

The short answer is that things written with a "denominator" of #0# are not numbers. Even #0/0# is not a number.

Explanation:

Any attempt to define #0/0 = a# for a number #a# 'breaks' arithmetic.

What this means is that, if you attempt to define #0/0# to be a number, that cannot be the only change you make in the number system.

You cannot say "Leave everything else the same, but make #0/0 = 1# or #0/0 = 0#."

Leaving everything else thet same will lead to a collapse of the number system.

Case 1

If #0/0 = a#, for #a != 0#,

then #2 * 0/0 = 2a#.

But #2*0/0 = (2*0)/0 = 0/0 = a#

So #2a=a#.

As another problem, what is #0/0+0/0#?

On the one hand, it shoulod be #a+a = 2a#, on the other hand it should be #0/0+0/0 = (0+0)/0 = 0/0 = a#

Dividing by #a# leads to #2 = 1# and arithmetic is broken.

Case 2

If #0/0 = 0#,

then

#1 = 0+1 = 0/0+1/1#

Get a common denominator using #a/b+c/d = (ad+bc)/(bd)# -- standard arithmetic

# = (0(1)+0(1))/(0(1)) = (0+0)/0 = 0/0 = 0#

The conclusion that #1 = 0# breaks arithmetic.