Is it possible to use mathematical induction to prove a statement concerning all real numbers, not necessarily just the integers? 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 "delta-x", and seeing whether (it is true for the value x) implies (it is true for the value x+"delta-x");  or is this flawed in some way. 
 A: 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 $.
A: 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.
A: 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 $y=x-n\delta$.
To sum up, induction-like 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.
