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I have the following proposition to prove:

For all $m \in\mathbb\ Z$, $m \cdot 0 = 0 = 0 \cdot m$

I can use the following axioms:

  1. commutativity
  2. associativity
  3. distributivity
  4. identity for addition ($0$)
  5. identity for multiplication ($1$)
  6. additive inverse
  7. cancellation: Let $m,n,p$ be integers. If $m \cdot n = m \cdot p$ and $m \ne 0$, then $n = p$.

Here is my proof:

\begin{align*} m \cdot 0 &= m \cdot (m + (-m))\\ m \cdot 0 &= (m \cdot m) + (m \cdot (-m))\\ m \cdot 0 &= (m \cdot m) +(m \cdot -1 \cdot m) \\ m \cdot 0 &= (m \cdot m) +-1 \cdot (m \cdot m) \\ m \cdot 0 &= (m \cdot m) - (m \cdot m) \\ m \cdot 0 &= 0 \end{align*}

However, I am not sure, given a simple set of axioms, that this solution is correct. More specifically, is factoring $-m$ as $-1 \cdot m$ acceptable? Or is another proposition that I should prove beforehand?

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  • $\begingroup$ Your work looks fine, but it completely depends on what you are allowed to assume (i.e., what axioms you are allowed to use). It would help if you stated what you are allowed to assume (existence of inverses, etc.). $\endgroup$ Commented Feb 6, 2015 at 20:03
  • $\begingroup$ @induktio Hi! I have added the axioms. :) $\endgroup$
    – Johnathan
    Commented Feb 6, 2015 at 20:09
  • $\begingroup$ I wouldn't use m. Just use 1 and -1. m*0 = m*(1 + (-1)). $\endgroup$
    – fleablood
    Commented Nov 7, 2015 at 8:14

2 Answers 2

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It depends. What are your axioms? What is your definition of the notation $-m$?

If $-m$ is defined as the additive inverse of $m$, then no, you cannot factor $-m = -1\cdot m$ until you prove that this is true.

EDIT: For the particular axioms you have listed, your proof in fact may well be circular, since the most straightforward way of proving that $-1\cdot m = -m$ is to add $m$ to $-1\cdot m$ and show that this sum is zero. Here's a hint to get you started for your original problem:

$$m\cdot 0 = m\cdot (0+0)$$ by the additive identity axiom. Can you see how to take it from here?

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  • $\begingroup$ Thank you! Yes, I have the inverse as axiom. $\endgroup$
    – Johnathan
    Commented Feb 6, 2015 at 20:12
  • $\begingroup$ Yes! I think I can. So, I would get: \begin{align*} m \cdot 0 &= m \cdot (0 + 0)\\ m \cdot 0 &= m \cdot 0 + m \cdot 0 \\ \end{align*} Then I can use a proposition that I have proven beforehand: Let $x \in\mathbb Z$. If $x$ has the property that for each integer $m$, $m + x = m$, then $x = 0$. In this case $m,x$ would both be $m \cdot 0$. Hence, $m \cdot 0 = 0$, no? $\endgroup$
    – Johnathan
    Commented Feb 6, 2015 at 20:30
  • $\begingroup$ No, that won't quite work, since here you have just $m\cdot 0$, not every integer $x$. What you can do, though, is add the additive inverse of $m\cdot 0$ to both sides, and then use associativity/additive inverse axiom/additive identity axiom a bunch of times. $\endgroup$
    – user7530
    Commented Feb 6, 2015 at 20:32
  • $\begingroup$ So, \begin{align*} m \cdot 0 &= m \cdot (0 + 0)\\ m \cdot 0 &= m \cdot 0 + m \cdot 0 \\ (m \cdot 0) + -(m \cdot 0) &= (m \cdot 0) + (m \cdot 0) + -(m \cdot 0)\\ 0 &= m \cdot 0 \end{align*} Then, I can use the symmetry and transitivity properties of = to state $m \cdot 0 = 0 = m \cdot 0$ Would that be correct? Thank you! $\endgroup$
    – Johnathan
    Commented Feb 6, 2015 at 20:40
  • $\begingroup$ You skipped a few steps, but yes that is the gist. You will need to use commutativity of multiplication to get $0 = 0\cdot m$. $\endgroup$
    – user7530
    Commented Feb 6, 2015 at 20:51
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Assume that m is an integer. By the commutative property we know that m.0 = 0.m.

Now, we only need to prove only that m.0 = 0. We use m = m, then

m.1 = m.1 because 1 is the identity under multiplication.

m.(1+0) = m.1 because 0 is the identity under addition.

Using the distributive property,

(m.1)+(m.0) = (m.1)

m +(m.0) = m

-m + m +(m.0) = -m + m (-m is the inverse of m under addition.)

(-m + m) +(m.0) = (-m + m), associative property.

0 +(m.0) = 0 because the definition of the identity under addition.

m.0 = 0 Q.E.D.

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  • $\begingroup$ @Johnathan It was a pleasure. $\endgroup$
    – Beginner
    Commented Feb 8, 2015 at 1:16
  • $\begingroup$ You skipped several steps at the end: associativity to get $(-m+m)+(m\cdot 0)$, then using additive inverses and additive identity to go from there to $m\cdot 0$. $\endgroup$
    – user7530
    Commented Feb 8, 2015 at 21:16
  • $\begingroup$ @user7530 Thanks for the advice! May I edit the answer using your ideas? $\endgroup$
    – Beginner
    Commented Feb 8, 2015 at 22:39
  • $\begingroup$ Yes, of course. $\endgroup$
    – user7530
    Commented Feb 8, 2015 at 22:44

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