I am referring to the part of proof by mathematical induction where you show that "if it is true for one value k then it is true for the value k+1". Does proof by induction work over all real numbers? I mean by considering any arbitrary change, say "deltax", and seeing whether (it is true for the value x) implies (it is true for the value x+"deltax"); or is this flawed in some way.

1The concept of induction asks for a wellordered set. That means that any nonempty subset contains a minimal element. – drhab May 30 '14 at 14:34

1Related: http://math.stackexchange.com/questions/450652/howcaniexpandmathematicalinductiontorationalnumbers – leonbloy May 30 '14 at 14:36
3 Answers
Yes. There are forms of induction suited to proving things for all real numbers. For example, if you can prove:
 There exists $a$ such that $P(a)$ is true
 Whenever $P(b)$ is true, then there exists $c > b$ such that $P(x)$ is true for all $x \in (b,c)$
 Whenever $P(x)$ is true for all $x \in (d,e)$, then $P(e)$ is true
then it follows that $P(x)$ is true for all $x \geq a $.

No, if the $c$s approach a limit, then by the third point $P(x)$ is true for that limit, and then by the second point $P(x)$ is true for some value above that limit. Thus, there can be no upper limit for which $P(x)$ is true. – user3294068 May 30 '14 at 15:04

Yes, I missed that you got the upper limit in the third point. It works fine. – Ross Millikan May 30 '14 at 15:05
I could conceive of a situation where one might have a statement (for example, an identity involving the floor function) for which you could have an induction part and a direct proof part. For instance, you could prove that a particular identity is true for all reals in $[0,1)$, and then extend that proof via induction over all intervals of the form $[k, k+1)$ for all integers $k$, thereby establishing the identity for all reals. But I don't think this is quite what you had in mind.
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1This would get all the positive reals. The same approach would work for the negatives, but it seems important to point out that the naturals have a minimum element and the reals do not. – Ross Millikan May 30 '14 at 14:43

1Yes, but suppose you could work both waysthe statement could be about an absolute value or a square as well, so if it works for positive reals, it also works for negative reals. I know it's sort of trivial but I'm just trying to establish plausibility rather than something mathematically interesting. – heropup May 30 '14 at 14:45

1I agree. I thought that there are several pieces that make induction work on the naturals and having a minimal element makes a natural starting place. It can be fixed, but there isn't such a natural place in the reals. – Ross Millikan May 30 '14 at 14:49
Let $A\subset\Bbb R$ and $\delta>0$. If $[0,\delta)\subset A$ and the statement $$x\in A\implies \{x\delta,x+\delta\}\subset A$$ is true for every $x\in\Bbb R$, then $A=\Bbb R$.
Indeed, let $y\in\Bbb R$. Suppose for now that $y\geq 0$. Let $n=\lfloor\frac y\delta\rfloor$. Then $y=n\delta+x$, where $x\in[0,\delta)$. The set $B=\{n\in\Bbb Z_{\geq 0}:n\delta+x\in A\}$ is inductive and contains $0$, so $B=\Bbb Z_{\geq 0}$ and $y\in\Bbb R$.
For $y<0$ just consider $n=\lfloor\frac {y}\delta\rfloor+1$ and write $y=xn\delta$.
To sum up, inductionlike arguments in $\Bbb R$ are valid if you prove your statement for all the numbers of some interval and that 'jumps' of the same length as the interval are guaranteed leftwards and rightwards.
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