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Is there a valid proof that x * 0 = 0? The definition of multiplication under the Peano axioms makes it axiomatic.

• Shamiq 17y ago
  I'm thinking something that has to do with set theory. I don't remember the steps.
• lmkg 17y ago
  It depends on what your axiom system is, what 0 is, and what your universe is. In Peano arithmetic it's axiomatic. Meanwhile, for fields (and possibly less-structred rings) in abstract algebra, it's a theorem. An important note is that the proof for fields requires negative numbers, while Peano arithmetic only describes the natural numbers. Here's the field proof I'm pulling out of memeory: x*0 = x*(y + -y) = x*y + x*(-y) = x*y - x*y = 0 This relies on the existence of additive inverses for the first step, distribution in the second step, and the fact that x(-y) = -(xy) in the third step. That third property can be derived from sufficiently-specific ring axioms, but I forget how specific they have to be. It might be true in any ring by virtue of distribution but I forget the proof. So to answer your question, yes there are... but it depends on the context. In some situations it may need to be axiomatic. A recursively-defined system like Peano may need to take it axiomatically as a base case.
  • warwick 17y ago
    Thank you. That's beautiful.
  • python123 17y ago
    0 is additive identity. 0+y=y for all integers y. 0 is an integer, so 0+0=0. 0+0 is closed under addition, so (0+0) is an integer, so x(0+0) = x0 for all integers x. By distributive law, x0 + x0 = x0. By closure under multiplication, x0 is an integer. By additive inverses, there exists an integer (-x0), such that x0 + (-x0) = 0. Because (-x0) is an integer, x0 + x0 + (-x0) = x0+(-x0). By associative property of addition and the transitive property of equality, x0 + 0 = 0. By the additive identity, x*0 = 0.
    • lmkg 17y ago
      Excellent walk-through, thank you. You also removed one of my uses of additive inverses, which gets us closer to not needing the negative integers. I have two ideas for getting rid of the last use of them in your last step, but I'm sure they're kosher. 1) From the equation x0 + x0 = x0, you don't need inverses, just cancellation. I believe that cancellation is a strictly weaker property. 2) From the equation x0 + x0 = x0, note that x0 is the additive identity. Inverses are unique, thus x0 = 0. Unfortunately, I'm not familiar enough with Peano arithmetic to know if proving either of these statements requires the statement we're trying to prove, that x*0 = 0. I'm more familiar with algebra, where inverses exist axiomatically. But at least we've weakened the hypotheses!
      • jules 17y ago
        To prove something about multiplication you first need to provide a definition of multiplication. So what is you definition of multiplication? The usual definition is: a*0 = 0 a*S(b) = a*b+a Cancellation follows from injectivity of S by induction.
      • python123 17y ago
        No, you can't say x0 is the additive identity because 0 is the only additive identity. That would be equivalent to x0=0. Your way is fine if you just use what archgoon wrote to complete it.
    • greenlblue 17y ago
      This is exactly the proof in the case of vector spaces.
  • archgoon 17y ago
    x(-y)=x(-y+y-y)=x(-y)+xy+x(-y) Cancellation yields 0=xy+x(-y) Therefore -(xy)=x(-y)
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• Jach 17y ago
  See: http://us.metamath.org/mpegif/mul01i.html http://us.metamath.org/mpegif/mul01i.html http://us.metamath.org/mpegif/mul02i.html http://us.metamath.org/mpegif/mul02i.html