Let f be a function from Z to R, such that f(x)=x/10, then f is
a) an increasing function
b) a strictly increasing function
c) a decreasing function
d) an onto function
Let "f" be a function from "\\mathbb Z" to "\\mathbb R", such that "f(x)=\\frac{x}{10}". If "x<y", then "\\frac{x}{10}<\\frac{y}{10}", and hence "f(x)<f(y)". It follows that "f" is a strictly increasing function.
The function "f" is not an onto function. Indeed, for "y=\\frac{1}{20}" the equation "f(x)=\\frac{1}{20}," which is equivalent to "\\frac{x}{10}=\\frac{1}{20}" and hence to "x=\\frac{1}{2}," has no solution in the set "\\mathbb Z" of integer numbers. Therefore, there is no integer number "x" such that "f(x)=\\frac{1}{20}."
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