what is the proper notation to make a note to yourself?

I know this is kind of silly but on exams we have to label the steps when using mathematical induction. For the induction step it really helps me if I start with "prove { [statement goes here] } " I'm loosing marks because technically {} are for sets, so what can I use?

Re: what is the proper notation to make a note to yourself?

Quote:

Originally Posted by

**Jskid** I know this is kind of silly but on exams we have to label the steps when using mathematical induction. For the induction step it really helps me if I start with "prove { [statement goes here] } " I'm loosing marks because technically {} are for sets, so what can I use?

That's harsh! Just write,

Step 1: Proving the base case.

Step 2: Proving that if holds for k holds for k+1

or something. Otherwise, take the marks on the chin, because there's nothing wrong with a bit of notational abuse! What you have written isn't ambiguous.

Alternatively, ask the person doing your marking what you should write. They'll know what they want...

(Was this an exam, or homework? I dock marks for silly things with my first-years, for example writing $\displaystyle 1\frac{1}{2}$ for your answer will get a big red cross through it - it should be $\displaystyle 1+\frac{1}{2}$, or $\displaystyle \frac{3}{2}$. But in a test I wouldn't care (big cross through it, yes, but they'd get the marks!)).

Re: what is the proper notation to make a note to yourself?

i usually use the following format:

proof of (statement, theorem, assertion, whatever) for n = 1:

(stuff)

assume (statement, theorem, assertion, whatEVER!) true for n = k ((by/as) induction hypothesis).

(stuff proving that true for k means true for k+1)

....therefore true for n = k+1, done, by induction on n.

*********

in your particular case, i would think you could just drop the brackets, and use the plain [statement goes here] whatever [statement] is (the brackets HERE

just being notational). like this:

prove by induction that the n-th square is the sum of the first n odd numbers.

proof:

the 1st square is the sum of the first odd number:

$\displaystyle 1^2 = 1$

suppose that $\displaystyle k^2 = \sum_{m=1}^k (2m-1)$ (the 2nd odd number is 3 = 2(2) - 1, the 3rd odd number is 5 = 2(3) - 1, so the k-th odd number is 2k-1).

then $\displaystyle (k+1)^2 = k^2 + 2k + 1 = \left(\sum_{m=1}^k (2m-1)\right) + 2k + 1$ (by our induction hypothesis)

$\displaystyle = \left(\sum_{m=1}^k (2m-1)\right) + 2k + 2 - 1$

$\displaystyle = \left(\sum_{m=1}^k (2m-1)\right) + 2(k+1) - 1= \sum_{m=1}^{k+1} (2m-1)$

therefore, by induction, for all (natural numbers) n, the n-th square is the sum of the first n odd numbers.