Proof using addition and multiplication axioms

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We start with $(k-m)+(m-n)$, now we start with the first step you mentioned:

1) $\big(k+(-1)m\big)+\big(m+(-1)n\big)$

For clarity, let's denote $m+(-1)n=a$, then we have $\big(k+(-1)m\big)+a$.

2) $\big(k+(-1)m\big)+a = k+\big((-1)m+a\big)$ by associativity.

Finally, we have $$k+\big((-1)m+a\big) = k+\big((-1)m+m+(-1)n\big) = k+\big(0+(-1)n\big) = k-n. $$

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Updated on August 01, 2022

Comments

  • user928112
    user928112 over 1 year

    I'm working on addition and multiplication axioms of integers for discrete math. I'm trying to prove (k - m) + (m - n) = k - n. The first step I took was this (without citing any of the axioms):

     = k + (-m) + m + (n)
    

    Is this incorrect?

    On a related note, the solution in the textbook on the other hand starts off like this:

     = (k + (-1)m) + (m + (-1)n)
     = k + [(-1)m + m + (-1)n)
    

    After that, (-1)m + m is zero. What I don't understand is how do they go from step 1 to step 2?

    The axioms I'm given are as follows: given axioms

    • user928112
      user928112 over 9 years
      @Berci Updated the question with the given axioms.
  • user928112
    user928112 over 9 years
    Smooth! I get it now. setting "m + (-1)n" to "a" does indeed make it easier to think about which step to take next.
  • diff_math
    diff_math over 9 years
    A good notation can be your friend...or your enemy, if it's bad notation :p