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It seems obvious that $$\forall n \in \Bbb Z^+, \sum_{i=1}^n{1} = n $$

However, I'm having trouble coming up with a formal proof for this.

Given a concrete number like $4$, we can say that $$\sum_{i=1}^4{1} = 1+1+1+1 = 2+1+1=3+1=4$$ by the definitions of summation, 2, 3, and 4. But how can we formally show this for an arbitrary positive integer?

I wonder if we can use the definitions of multiplication and the multiplicative identity, but I'm not sure.

Daniel W. Farlow
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Ypnypn
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1 Answers1

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Hint: Use induction.

To get you started, suppose we have the statement $$ S(n) : \sum_{i=1}^n 1 = n $$ that you are trying to prove. Fix some $k\geq 1$ and assume $$ S(k) : \sum_{i=1}^k 1 = k $$ is true. Then we need to show that $$ S(k+1) : \sum_{i=1}^{k+1} 1 = k+1 $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \sum_{i=1}^{k+1} 1 &= \sum_{i=1}^k 1 + 1\tag{using definition of $\Sigma$}\\[1em] &= k + 1\tag{by $S(k)$}, \end{align} we reach the right side of $S(k+1)$.

The result, then, follows by mathematical induction.

Daniel W. Farlow
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    Don't skip the case $k=1$. Given your name, I'd assume you wouldn't forget that. The above proof without $k=1$ would also show $\sum_{k=1}^n 1 = n+1$. :) – Thomas Andrews Jan 26 '15 at 22:08
  • @ThomasAndrews Absolutely--never forget the base case! I just wanted to get the OP started though. I did not want to do the entire proof for him. Also, what I provided, as we both know, is simply the mechanics of the main part of the proof. If this were to be turned in, then I would expect a much more fluid composition as I have provided in some other induction proofs. Good point though :) – Daniel W. Farlow Jan 26 '15 at 22:10