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#REDIRECT[[mathematical induction]]
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This article presumes familiarity with [[mathematical induction]]. That article, not this one, is the appropriate source of information for those not so familiar.
==The three forms==
Proofs that a subset of { 1, 2, 3, ... } is in fact the whole set { 1, 2, 3, ... } by [[mathematical induction]] usually have one of the following three forms. They are collected here in order to show the contrast between them, and its coexistence with the commonality between them.
===First form===
* The basis for induction is trivial;
* the substantial part of the proof goes from case ''n'' to case ''n'' + 1.
===Second form===
* The ''first'' case is [[vacuous truth|vacuously true]];
* the step that goes from the ''n''th case to the (''n'' + 1)th case is trivial if ''n'' ≥ 2 and impossible if ''n'' = 1;
* the substantial part of the proof is the ''second case'', and the second case;
* the second case is relied on in the trivial induction step.
===Third form===
* The induction step shows that if ''P''(''k'') is true for all ''k'' < ''n'' then ''P''(''n'') is true (proof by [[complete induction]]);
* no basis for induction is needed because the first, or basic, case is a vacuously true special case of what is proved in the induction step.
This form works not only when the values of ''k'' and ''n'' are natural numbers, but also for ordinal numbers; see [[transfinite induction]].
==Examples==
This is not the appropriate place to go into details of proofs, but rather only broad outlines of proofs are shown. For examples of proofs by mathematical induction with full details, see [[mathematical induction]].
===First form===
Most proofs by mathematical induction that are adduced as examples when mathematical induction is first taught are of the first form.
===Second form===
====Product rule====
The usual [[product rule]] taught in [[calculus]] says
:<math>(fg)' = f'g + g'f.\,</math>
For a product of ''n'' functions, one has
:<math>(f_1 f_2 f_3 \cdots f_n)' \,</math>
::<math>= (f_1' f_2 f_3 \cdots f_n)
+ (f_1 f_2' f_3 \cdots f_n)
+ (f_1 f_2 f_3'\cdots f_n)+\cdots +(f_1 f_2 \cdots f_{n-1} f_n').\, </math>
In each of the ''n'' terms, just one of the factors is a derivative; the others are not.
Now observe four facts:
*If ''n'' = 1, then the proposition just says
::<math>f' = f'.\,</math>
:It can be said that just one factor is a derivative, and it is [[vacuous truth|vacuously true]] that the others are not, because there are not others!
*To go from case ''n'' to case ''n'' + 1 is trivial if ''n'' ≥ 2, but impossible if ''n'' = 1.
*To go from case ''n'' to case ''n'' + 1, one must ''use'' the case ''n'' = 2.
*The substantial part of the proof is the case ''n'' = 2, and that is precisely the product rule that is taught differential calculus.
====Triangle inequality====
The usual [[triangle inequality]] in [[metric space]]s says
:<math>d(a,c) \le d(a,b) + d(b,c).\,</math>
For a sequence of ''n'' points, one has
:<math>d(x_1,x_n) \le d(x_1,x_2)+\cdots+d(x_{n-1},x_n).\,</math>
[More examples to be added.]
[[Category:Mathematical logic]][[Category:Proofs]]
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