# How can I show that by induction? [closed]

I have asked that question - Dividing square of 2013x2013 - and I receive useful answer, which was correct, of course, but because solution of that exercise is my homework I have to have an accurate (best step-by-step) solution. user2566092 has written that I should use to solve that problem complete math induction, but I totally don't know how to do that ! I read an article in Wikipedia, Math.com, etc. , etc. but I don't know how induction can be useful in my problem. Please give me formulas and solutions, because I have to do that on Monday (please, not hints, I never used induction and don't know how it can be helpful for me - I know only general rules ). Thank you in advance ! Please give me fast accurate solution.

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## closed as off-topic by Antonio Vargas, PVAL, drhab, M Turgeon, Daniel FischerAug 3 at 18:24

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So you're asking for someone to produce a solution for you to copy directly with minimal understanding? –  Calvin Lin Oct 12 '13 at 15:54
@CalvinLin I don't know how to use complete induction, I know only basic rules of induction and I am asking how to do that and for solution - everything is ok - in FAQ there is written like that –  user99183 Oct 12 '13 at 15:55
See math.stackexchange.com/questions/17041/… and look for Arthuro's linked answer in one of the comment –  Jean-Sébastien Oct 12 '13 at 16:05
Help me please ! –  user99183 Oct 17 '13 at 15:40

Define the row in which a vertical domino starts at to be the lowest row in which it appears in.

Consider the first row. Show that the number of vertical dominos in the first row is (fill in the blank). These dominos must start in the (fill in the blank) row(s). Hence, the number of vertical dominos that start in the first row is (fill in the blank).

Consider the second row. Show that the number of vertical dominos in the second row is a (fill in the blank). These dominos must start in the (fill in the blank) row(s). Hence, the number of vertical dominos that start in the second row is (fill in the blank).

Consider the third row. Show that the number of vertical dominos in the third row is a (fill in the blank). These dominos must start in the (fill in the blank) row(s). Hence, the number of vertical dominos that start in the third row is (fill in the blank).

Continue for all 2013 rows ... (Fill in the blank).

Hence, the number of vertical dominos is (fill in the blank).

You do not need to use the language of induction, and can express it simply as above. When you learn the technique of strong induction, revisit how to simplify this proof (and even then, not by much).

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what is (fill in the blank) ? –  user99183 Oct 12 '13 at 16:02
@Ty221 It is your homework. –  Calvin Lin Oct 12 '13 at 16:02
but I am asking you for an answer –  user99183 Oct 12 '13 at 16:04
@Ty221 My answer, is for you to do your own homework, with guidance. The above is very easy to fill in the blanks, esp if you understood what was happening in your other question. –  Calvin Lin Oct 12 '13 at 16:05
You should not accept an answer that you don't understand (and this applies to my answer to). You could have asked @user2566092 to explain his answer in more detail, though I believe that there is enough written out for you to figure it. I do not believe in spoon-feeding / assisting you to cheat on your homework. –  Calvin Lin Oct 12 '13 at 16:09

To prove $p_n$:

Show that $p_1$ is true. (This is the base case.)

Show that $p_{k+1}$ is true, given $p_k$. (This is usually the tricky step. Now, since we have shown $p$ is true for { 1, 1+1, 1+1+1, ... }, we can say that $p_m$ is true for any $m \in \mathbb N$.)

Therefore, $p_n$ is true. ($n \in \mathbb N$)

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