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Show that f:r->r defined by f(x)=1/x is one-one and onto, where R? is the set of all non-zero real numbers. Is the result true, if the domain R? is replaced by N with co-domain being same as R

Given f:R??R? defined by f(x)=1 / x where R? is a set of nonzero real numbers:
Let x and y be two elements in R?.
Step1: Injective or One-One function:

For a one-one function, f(x)=f(y)
?1 / x=1 / y?x=y.
Therefore f:R??R? defined by f(x)=1 / x is one-one.

For an on-to function, for every y?Y, there exists an element x in X such that f(x)=y.

? For every y?R? there must exist x=1 / y?R? such that f(x)=1 / (1/y)=y.
Therefore f:R??R? defined by f(x)=1x is onto.

Step 2: Surjective or On-to function:

Now, let us consider a function f2 : N?R? defined by f2(x)=1/x:
For a one-one function, f2(x2)=f2(y2)
?1 / x2=1 / y2? x= y2.
Therefore, f2: N?R? defined by f2(x)=1 / x is one-one.

? For and onto function, for every y2?R? there must exist x2=1/y2?R? such that f2(x)=1/(1/y2)=y2.
However, we see that for y2=1.5?R?, there is no x2 in N such that f2(x)=1/1.5.
Therefore  the function f2 is not onto.

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