Proof using addition and multiplication axioms
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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user928112
Updated on August 01, 2022Comments
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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:
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user928112 over 9 years@Berci Updated the question with the given axioms.
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user928112 over 9 yearsSmooth! I get it now. setting "m + (-1)n" to "a" does indeed make it easier to think about which step to take next.
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diff_math over 9 yearsA good notation can be your friend...or your enemy, if it's bad notation :p