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Proofs by Induction

No discussion of recursion and induction would be complete without some obligatory examples of proofs using induction. We start with the Hello World example.

Proofs by Induction

No discussion of recursion and induction would be complete without some obligatory examples of proofs using induction. We start with the Hello World example.

For our first version of a proof of , we clearly identify the open statement \(S_n\) and describe the proof carefully in terms of \(S_n\). As you develop more experience with writing proofs by induction, this will become less essential, as you'll see in the second version of the proof.

Proof

Let \(n\) be a positive integer, and let \(S_n\) be the open statement \[\begin{aligned}\end{aligned}\] We will prove that \(S_n\) is true for all positive integers by induction. For the basis step, we must prove that \(S_1\) is true. When \(n=1\), the left-hand side of \(S_n\) is just \(1\), while the right-hand side evaluates to \(1(1+1)/2=1\). Therefore, \(S_1\) is true.

Next we assume that for some positive integer \(k\), \(S_k\) is true. That is, we assume \[\begin{aligned}\end{aligned}\] We now seek to prove that \(S_{k+1}\) is true, and begin by considering the left-hand side of \(S_{k+1}\). We notice that \[\begin{aligned}\end{aligned}\] since our inductive hypothesis that \(S_k\) is true gives us the simpler formula for the summation. Now continuing with a bit of algebra, we find \[\begin{aligned}\end{aligned}\] Therefore, \(S_{k+1}\) is true. Since we have shown that \(S_1\) is true and that for every positive integer \(k\), if \(S_k\) is true, then \(S_{k+1}\) is true, we conclude that \(S_n\) is true for all positive integers \(n\) by the .

Before looking at a refined version of this proof, let's take a moment to discuss the key steps in every proof by induction. The first step is the basis step, in which the open statement \(S_1\) is shown to be true. (It's worth noting that there's nothing special about \(1\) here. If we want to prove only that \(S_n\) is true for all integers \(n\geq 5\), then proving that \(S_5\) is true is our basis step.) When proving the basis step, if \(S_n\) is an equation, we do not just write down \(S_1\) and move on. We need to prove that \(S_1\) is true. Notice how in the proof above, we discussed the left-hand side of \(S_1\) and the right-hand side of \(S_1\) and concluded that they were equal.

Now let's take a look at a more refined proof of . From here on, when we give a proof by induction, we'll use this style. As you're getting started with induction proofs, you may find it useful to be more explicit about the steps as we did in the first proof above.

Condensed: the full section is in Keller & Trotter, Applied Combinatorics.

Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.

Tababar (2)

Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.

  1. Suppose that \(t(n)\) is a recursive function defined for \(n\geq 1\) such that \(t(1)=1\) and for \(n\geq 2\), \(t(n)=2t(n-1)+1\). Create a proof that for all \(n\geq 1\), \(t(n) = 2^n - 1\).

  2. Write a couple of sentences explaining why the making the inductive hypothesis is not the same as assuming what we are trying to prove.

Ku hay shaqadaada

A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.

Ku soo biir Ku soo gal

Calaamadaha la isticmaalay halkan

Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Su'aalaha dadku waydiiyaan

Permutation or combination?

Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).

What is a graph in this sense?

Dots (vertices) joined by lines (edges), not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.

Qaybaha ee boggaan waxaa la beddelay ka Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo.

In ka badan Combinatorics & Graph Theory