-ve number_of_times ?

  • Thread starter jobyts
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  • #1
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-ve number_of_times ???

Let's say I'm trying to teach a kindergartener about addition/subtraction/multiplication of positive/negative numbers.

Let' say she has 5 candies. I can ask her - how many candies you 'have'? The answer is +5.
Or, if she has no candies and 'needs' 5 candies, that could be represented as -5.

If one kid needs 5 candy and another kid needs 3 candy, totally they need 8 candy,
hence (-5) + (-3) = (-8) => -ve number + -ve number = -ve number

If one kid needs 5 candy and there are 3 such kids, in total, they need 15 candies, hence
(-5) * (+3) = (-15) => -ve number * positive number of times = -ve number.

The question I have is how do I say the candy example for multiplication of 2 -ve numbers.

How's multiplication of 2 negative numbers, makes into a positive number?

Shouldn't '-ve number_of_times' be considered as an 'undefined' operation, like divide_by_zero? Did we just define a meaningless thing, and to make things work, we further kept on defining more fancy things like complex numbers.., real and imaginary part...
 

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  • #2
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if you insist on that example, if each kid need 5 candy and we have 3 such confused kids, then they actually have 15 candies. Here a "confused kid" is a kid who thinks he need but actually have a given number of candies, and vice verse!
 
  • #3
Office_Shredder
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Shouldn't '-ve number_of_times' be considered as an 'undefined' operation, like divide_by_zero?

a negative number times a negative number is positive because: given a, -a is defined to be the unique number such that a + (-a) = 0

Given a positive number a, -a is a number. Hence there exists -(-a) such that -(-a) + (-a)=0

But a+(-a) = 0. As -(-a) is the unique number with this property, we deduce -(-a) = a

And this is why two negatives gives a positive.

The problem with all your examples that you're trying is that the number of kids is a natural number, not an integer, so you're struggling to find an example of a negative number of kids. In fact, there are very few basic examples where you multiply two things together and neither of them is naturally 'counted'

EDIT: Here are a couple of good ones:
http://mathforum.org/dr.math/faq/faq.negxneg.html
 
  • #4
JasonRox
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Wow! That has to be the worth method ever. Damn.

First, get the child curious. That gives motivation.

Second, get a new lesson.
 
  • #6
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Thanks for all the answers.

Another question: Is there a proof that a number should have an imaginary part, or it is a thin airconcept?
 
  • #7
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Every COMPLEX number has both a real and imaginary part. There is no proof, that is the definition of a complex number.
 
  • #8
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Every COMPLEX number has both a real and imaginary part. There is no proof, that is the definition of a complex number.

Is there a proof for complex number to exist? Can I just make a statement the imaginary part of a complex number can have a real part and imaginary' part and goes on infinitely recursively. What makes the maths world to accept the definition of complex numbers to be the current one, and reject all others? Is it the applications of it makes a definition standard?
 
  • #9
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I'm not sure what "proof of existence" do you seek here, would you agree that the real numbers exist?
I'll restrict myself to the mathematics though, leaving existence and usefulness in waves, quantum mechanics, ... to physicists!

Given any field R, there exist a field C which contains R as a sub-field, this "larger" field is unique (up to k-isomorphism), and every non-constant polynomial over C (and consequently over R) has a root that belong to C.
Or, in other words, there's a unique(up to k-isomorphism) algebraically closed field extension for every field.

So that when we consider the real numbers, we know for sure (without construction) that a larger field must exists which is algebraically closed and is unique. (But thanks to the fundamental theorem of algebra, we know that the structure of the complex numbers meets these criteria)

Moreover, we are not restricted to single construction, as long as we have a field which has the reals R in it (as a subfield), and contains all roots of R.

Here's another construction of C which doesn't require the usual "adjoining i". Consider the quotient of the set of all polynomials in X over the real numbers by the ideal of the irreducible polynomial [itex](X^2 + 1)[/itex].
 
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