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.
Is it possible to use mathematical induction to prove a statement concerning all real numbers, not necessarily just the integers?
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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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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.
Yes. There are forms of induction suited to proving things for all real numbers. For example, if you can prove:
then it follows that $P(x)$ is true for all $x \geq a $.