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If we have a number $x$ such that $0 \leq x \leq 1,$ using only addition, subtraction, multiplication and division ops can we go from $x$ to $1$, without knowing what the original $x$ value is?

Example: Let $x = 0.5,$ we can reach $1$ by doing $2 * x.$ However, this only works for $x = 0.5$ and not other possible $x$ values such as $0.2, 0.35$ or $0.87.$

A basic solution is dividing $x$ by itself. Are there any other functions we can use such that $f(x) = 1?$

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    $\begingroup$ The function "dividing by itself" is just the function $f(x)=1$ for $x>0.$ A function in not concerned with the way it is computed, only with the result. $f(x)=1$ is the most basic such function. $\endgroup$ – Thomas Andrews Sep 3 '19 at 20:42
  • $\begingroup$ The ceiling function $f(x)=\lceil x\rceil$ works for the interval $0\lt x\leq 1$. $\endgroup$ – Andrew Chin Sep 3 '19 at 20:46
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    $\begingroup$ Maybe $f(x)=x+(1-x)$ ? $\endgroup$ – Fareed Abi Farraj Sep 3 '19 at 20:51
  • $\begingroup$ There are infinitely many: $x-x+1$ or $2x-2x+1$ or $3x-3x+1$ or ..., but I have the feeling that this kind of answer is not satisfying for you. $\endgroup$ – M. Winter Sep 3 '19 at 21:46
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    $\begingroup$ @FareedAF ... of course if you already know $1$ you are done, so there would be no need to do $x+(1-x)$. $\endgroup$ – GEdgar Sep 3 '19 at 21:48
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What if $x=0$? Then addition, subtraction, and multiplication can still only yield $0$ again. And division by $0$ is undefined. Therefore: starting with $x=0$ and using only addition, subtraction, multiplication, and division never yields the answer $1$.

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  • $\begingroup$ Why not $0+1=1$ ? $\endgroup$ – Fareed Abi Farraj Sep 3 '19 at 21:51
  • $\begingroup$ If I do not know "1", then I cannot do $0+1$ of course. $\endgroup$ – GEdgar Sep 4 '19 at 0:06
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It seems you did not recognize that multiplying $0,5×2$ you are also dividing $\frac{x}{x}$ since multiplying by $2$ is the same as dividing bei $1/2$, so in fact you have with your operations only $\frac{x}{x}$ so with $0,35$ you multiply by $\frac{100}{35}$.

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