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Junseong,

Today at the end of class, I don't think we fully communicated. I think it's important that we do understand each other on this topic, because I'm pretty sure that the answer you gave would NOT be given credit on a Master's qualifying exam. It's not the algebra that's the problem, rather it is logical implications (and your lack of correctly spelling out how each statement is logically connected to the other) that caused the problem.

You had something like the following on your exam.

  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2} \text{(adding $k+1$ to both sides}.$$ This shows that $S_{k+1}$ is true.

From what you wrote above, it is impossible for me to determine, without guessing, what your logical implications are. You have not fully communicated what you wanted to say. Which of the following did you mean?

  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ Adding $k+1$ to both sides of the above and simplify gives $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}.$$ This shows that $S_{k+1}$ is true.
From Ben: The above is 100% correct, and clearly shows the order of all logical implications. You did not include the sentence in the middle of your computation, rather you left a parenthetical comment after your work that makes your reader guess your meaning. All of the example that follow are ways that your original statement could have been interpreted. You should see in each case that something is wrong.
  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ We need to prove that $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}$$ is true. Adding $k+1$ to both sides proves that $S_{k+1}$ is true. Adding $k+1$ to both sides of what ($S_k$ or $S_{k+1}$)? Also, adding $k+1$ doesn't give that result. Did you do some simplifications afterwards? State that you did.
  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ We then know that $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}$$ is true, because we can adding $k+1$ to both sides. This proves that $S_{k+1}$ is true. Did the student say "add" when they actually meant "substitute"? So they really meant "replacing $k$ with $k+1$". This is just flat out false. But without saying "and then simplifying the right hand side", you've left your reader guessing what you meant.)
  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ Hence we know $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}.$$ Somehow adding $k+1$ to the left side makes this true (not sure what happened to the right side). This shows that $S_{k+1}$ is true. Did the student not really know how to obtain $S_{k+1}$ from $S_k$, and they just said $S_{k+1}$ is true together with noting that $k+1$ was added to the left.
  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ We need to show $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}.$$ Somehow adding $k+1$ to both sides of the first equation makes the second equation true (not sure how, but it does). This shows that $S_{k+1}$ is true.
  • For some $k\in\mathbb{N}$ assume that $S_k$ is true. This means that $$1+2+\cdots +k=\frac{k(k+1)}{2}.$$ We need to show $$1+2+\cdots +k+k+1=\frac{(k+1)(k+2)}{2}.$$ Somehow adding $k+1$ to both sides of the second equation makes $S_{k+1}$ true (not sure how, but it does).