I faced this problem once:
Let's say we have given even integer $n$, now we want to prove that there isn't set of odd numbers $S$ such that $S_{1} + S_{2} + \dots + S_{k} = n \text { and } k \text{ is odd number} $.
My thinking
We know that $S = \{S_{1}, S_{2}, \dots, S_{k}\}$, and we have that $k\text{ mod } 2 = 1$. If we set each element in the set $S$ to have the value $S_{i} = S_{i} \text{ mod } 2$, for each $1 \leq i \leq k$. Because the set $S$ consists only odd numbers it will have the form $S = \{1, 1, \dots , 1\}$, or there will be $k$ ones. Because $k$ is odd if we sum the values they will have odd value, so it is impossible to write even number $n$ of odd number of odd numbers.
Answer
I'd say your proof is correct and sufficient, unless you are a beginner in math. In that case, there are some technically vague parts of the proof you should improve, since they are a litle hand wavy.
For example, the sentence
If we set each element in the set $S$ to have the value $S_i=S_i\mod 2$
isn't really mathematically rigorous. Technially speaking, if you do that, you are left with the set $\{1\}$ and you can't prove anything from that. Now, sure, I know what you meant, but if you are studying math for 2 years or less, then "you know what I meant" just isn't good enough. Rigorous mathematical language is vital in math and you need to learn it.
So, if you are a "beginner" in math, I suggest you rewrite your proof without any vague hand-wavyness. You will probably notice that you will have to use induction to get anywhere, and that's fine. Induction is probably the easiest (and, arguably, the only) way to prove the statement rigorously.
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