Is the following an acceptable proof (be harsh (Nod))?

I mean, is the induction also required? Even if not, I need help with it:

- Base case - : ; define .

- Induction - : suppose .

Then ... how do I use the supposition? (Blush)

- Oct 1st 2010, 01:24 AMcourteousProve: each square is the sum of 2 consecutive triangular numbers
Is the following an acceptable proof (be harsh (Nod))?

I mean, is the induction also required? Even if not, I need help with it:

- Base case - : ; define .

- Induction - : suppose .

Then ... how do I use the supposition? (Blush)

- Oct 1st 2010, 01:50 AMProve It
Just show that

...

. - Oct 1st 2010, 02:16 AMcourteous
Done that, just didn't write it down. (Wink) The Halmos sign (with a "?" mark) indicates whether this is enough ... or is induction also needed (even if it is not, still help me with the induction step (Blush)).

MOD request: Can you please restore my spam-hijacked question? - Oct 1st 2010, 02:21 AMProve It

. - Oct 1st 2010, 02:33 AMcourteous
Is this induction (not saying that it isn't, I just don't know)? I mean, isn't it necessary to use equality?

- Oct 2nd 2010, 10:51 AMBruno J.
- Oct 2nd 2010, 12:13 PMcourteous
This is just a special case of . (Smirk)

Guess I learned something about proving things ... and not trying to prove the obvious. (Happy) - Oct 2nd 2010, 02:13 PMArchie Meade
You have given a very nice proof...

Suppose you were unaware of the proof you gave in your first post and were feeling lazy...

Here is the "Proof By Induction" method

The triangular numbers are 1, 3, 6, 10, 15, 21, 28,....

1=1^2

1+3=4=2^2

3+6=9=3^2 etc

**P(k)**

**P(k+1)**

Try to show that P(k) being valid (even if we don't know whether it is or not)

will**cause**P(k+1) to be valid.

Hence write P(k+1) in terms of P(k).

**Proof**

if P(k) really is valid

Now the line of dominoes is in place, hence you need to check if the first one falls.

true

true, depending on where you want to start from.