A few days back I was just going through the posts by different enthusiasts and found the following alleged fallacy:

$x\in N(the\quad set\quad of\quad naturals)\\ { x }^{ 2 }={ x }^{ 2 }\\ \Longrightarrow { x }^{ 2 }=x+x+x+..(x\quad times)\\ \Longrightarrow 2x=1+1+1+..(x\quad times)\quad (differentiating\quad both\quad sides)\\ \Longrightarrow 2x=x\\ \Longrightarrow 2=1$ Let me assure the reader that this is just a trivial fallacy. The fallacy can be solved by the following simple argument:

When one is defining f(x)=x^2=x+x+..+x(x times), one is tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) is not differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point (or limit point) of the set N, as the reader can very easily verify by taking any arbitrary natural number and choosing a neighbourhood of radius 1/2. Here lies the fallacy.

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## Comments

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TopNewestI think it might be easier to just prove that f(x) is not continuous (as it is only defined for non-negative integers) and hence it is not differentiable at any point. To analytically show that f(x) is not continuous you need to show that $\ lim_{x \rightarrow\ a} f(x) \neq f(a)$ which is true as x is not defined $\ \forall x \in (a-1,a+1) \ ,\ (i.e. a-1 < x < a+1) \ ....... \forall a \in Z$

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Curtis , your attempt is worthy of appreciation but let me make a point here that the $f(x)$ i have defined here is actually continuous throughout its domain(though it may seem weird since the commonplace notion of continuous functions makes the allusion of those functions which can be graphically presented "without lifting the pen from the paper", though this one is a series of dots on the graph plane).Let me state here the formal definition of a continuous function. Let $A\subseteq R$ and $f:A\rightarrow R$ and let $c\in A$. Then $f$ is said to be continuous at $c$ if given an $\varepsilon >0$ there exists a $\delta >0$ such that if $x\in A$ and $\left| x-c \right| <\delta$, then $\left| f(x)-f(c) \right| <\varepsilon$. If $f$ is not continuous at $c$, then it is said to be discontinuous at $c$. Now note that the definition does not make any mention of whether $c$ is a limit point of $A$ or not (I hope you know what are limit points or cluster points, they are just points such that any arbitrarily small neighborhood of them contains infinitely many points in $A$). If $c$ is so, then we have to check the three following conditions to testify the continuity of $f$ at $c$:(i) $f$ must be defined at $c$, so that $f(c)$ is meaningful.(ii) The limit of $f$ at $c$ must exist in $R$.(iii) The limit value must equal the functional value at $c$ . However, if $c$ is not a limit point of $A$, we see, there exists a neighborhood ${ V }_{ \delta }\left( c \right)$, of $c$, such that $A\cap { V }_{ \delta }\left( c \right) =\left\{ c \right\} \quad$. Thus we conclude that $f$ is by default continuous at those points in its domain $A$, which are not limit points of $A$, i.e., which are "isolated points" of $A$. Note that the domain of the function defined in the above note is $N$, the set of naturals and every point in $N$ (i.e., every natural number) is an isolated point in $N$. So our function is in fact automatically continuous at every point of its domain. So from now onward, whenever you are to check the continuity of a function at a point in its domain, always study whether the point is a limit point or an isolated point of its domain. Use the notion of "drawing the curve without lifting the pen at that point", only after you find the point to be a limit point. If the point is not so, then declare the function to be continuous at that point peremptorily. I hope this note helps you..:-)

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