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Why does multiplying by 0 equal 0?

  1. Dec 18, 2008 #1
    I know why it would equal to 0 if it was (0*0). But what about an actual number? Why does (100*0) equal to 0? You're not multiplying anything, but shouldn't it still equal to 100? If I have 100 cookies on the table, and I don't multiply it by anything, why do I suddenly have zero cookies on the table?

    I'm just trying to gain a conceptual understanding behind the zero-factor algebraic property.
     
  2. jcsd
  3. Dec 18, 2008 #2
    One possible conceptual way to think about it is to say, multiplying 2 * 100 is like having 2 groups of 100 cookies on the table. Multiplying 1 * 100 is like having 1 group of 100 cookies on the table. Multiplying 0 * 100 is like having no groups of 100 cookies on the table, hence no cookies at all.

    However, I agree that this is may not be very intuitive. When dealing with 0, it is sometimes more difficult to match mathematical situations to real life situations. It is probably better to understand 0 * any number = 0 just as a consequence of several properties of numbers that we take for granted.
     
  4. Dec 18, 2008 #3

    statdad

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    Remember that [tex] 0 + a = a [/tex], [tex] 0 + 0 = 0 [/tex], and [tex] a - a = 0 [/tex], no matter what number you use for [tex] a [/tex]. Now the rules of arithmetic give this

    [tex]
    \begin{align*}
    0 \cdot a & = (0 + 0)a\\
    & = 0 \cdot a + 0 \cdot a\\
    \left(0 \cdot a - 0 \cdot a\right) & = 0 \cdot a\\
    0 & = 0 \cdot a
    \end{align*}
    [/tex]
     
  5. Dec 18, 2008 #4

    Vid

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    You actually prove this at the beginning of an undergrad analysis/theoretical calculus class.
     
  6. Dec 18, 2008 #5
    That's actually a very good way of thinking about it.

    Ah, so it's sort of a cancelation?
     
  7. Dec 18, 2008 #6

    tiny-tim

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    Hi LightbulbSun! :smile:

    No, if you don't multiply it by anything, you still have 100 cookies on the table.

    This is a language thing

    "not multiplying by anything" is not the same as "multiplying by nothing" …

    "not multiplying by anything" means leaving it the same. :smile:

    EDIT: I think the French don't have this problem …
    they distinguish between (pardon my French! :smile:) …
    "multiplier par rien" and "ne multiplier par rien"
     
    Last edited: Dec 18, 2008
  8. Dec 18, 2008 #7

    symbolipoint

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    This is far too natural to be confusing. Remember that multiplication is repeated addition. If you add 100, ZERO times, you have ZERO. If you want 100 as result, then you must add 100 ONE time.
     
  9. Dec 18, 2008 #8
    Now it makes more sense to me. Thanks for the explanation.
     
  10. Dec 18, 2008 #9

    Hurkyl

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    The problem is that natural language sort of pigeon-holes us into thinking "none" and "at least one" are conceptually different. As soon as you break through this barrier and become comfortable working with degenerate cases, stuff like this becomes easy.

    For example, how many pennies do you have, if you have zero rows of N pennies each? (Or, as one would generally say in natural language, if you don't have any rows of N pennies each)
     
  11. Dec 18, 2008 #10
    proof ox=o

    0x = 0x + 0 = 0x + [ x + (-x)] = (0x + x) + (-x) = x( 0 +1) + (-x) = x +(-x) = 0

    or

    0x = 0 <===> 0x + x = 0+x <=====> x( 0 + 1) = 0 + x <===> x = 0 + x <===> x=x correct so 0x=0
     
  12. Dec 19, 2008 #11

    symbolipoint

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    How does a person learn to work with such simple, low-level ideas like that to prove what would otherwise seem so natural? Fantastic!
     
  13. Dec 22, 2008 #12
    Because it turns out simple and useful.

    Multiplication, and any operation, is something defined by the mathematician. We could imagine a world where 0 * n = n. But it would break a lot of useful theorems. For instance, 0 * 1 + 1= 1 + 1 = 2. However, 0 * 1 + 1 = (0 + 1)*1 (by distribution), so 0 * 1 + 1 = 1, and so 1 = 2. We must conclude that addition no longer distributes over multiplication (disastrous!!).

    In many definitions, when you get to the lowest possible value, the definition loses its literal intuitive meaning. One example is the factorial function, where 0! = 1. Factorial is often defined as the product: 1 * 2 * ... * n, but when n = 0, this definition doesn't make sense.

    (Though there are other definitions that do, this is just one example).
     
  14. Dec 22, 2008 #13
    There isn't really any reason why, that's the way it's defined.
     
  15. Dec 22, 2008 #14
    We have the winner.

    Mathematics is a human convention. That's how zero is defined in it.
     
  16. Dec 22, 2008 #15
    add it to what...anything?
     
  17. Dec 22, 2008 #16

    DaveC426913

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    Yes, anything.

    126,
    12,378,726,387,
    0,
    or simply n.



    i.e.:

    126 + 100*0 = 126
    126 + 100*1 = 126 + 100

    12,378,726,387 + 100*0 = 12,378,726,387
    12,378,726,387 + 100*1 = 12,378,726,387 + 100

    0 + 100*0 = 0
    0 + 100*1 = 0 + 100

    or simply

    n + 100*0 = n
    n + 100*1 = n + 100
     
  18. Dec 23, 2008 #17
    So...that would mean such is true for all systems?
     
  19. Dec 23, 2008 #18

    DaveC426913

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    I'm not prepared to say that. I am simply clearing up the possible confusion of the question "if you 'add' 100 times 0, what are you adding it to?".
     
  20. Dec 24, 2008 #19
    If by 0, you mean the element of the system such that a + 0 = a for all elements a in the system, then yes, as the property follows directly from this statement and the distributive property of multiplication over addition, and multiplicative commutativity. Hint: Consider the equation a*a= (a + 0)(a + 0).
     
  21. Dec 24, 2008 #20

    arildno

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    Change that to an identity..
     
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