Let R0 be the set of all non zero real numbers. Then, show that the fu

1.	Let R0 be the set of all non zero real numbers. Then, show that the function \(f:R_0→R_0:f(x)=\frac{1}{x}\) is one-one and onto.
Answer» To prove: function is one-one and onto Given: \(f:R_0→R_0:f(x)=\frac{1}{x}\) We have, \(f(x)=\frac{1}{x}\) For, f(x₁) = f(x₂) \(\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂ When, f(x₁) = f(x₂) then x₁ = x₂ ∴ f(x) is one-one \(f(x)=\frac{1}{x}\) Let f(x) = y such that \(y\in R_0\) \(\Rightarrow y=\frac{1}{x}\) \(\Rightarrow x=\frac{1}{y}\) Since \(y\in R_0\) \(\Rightarrow \frac{1}{y}\in R_0\) ⇒ x will also \(\in R_0\), which means that every value of y is associated with some x ∴ f(x) is onto Hence Proved

Let R0 be the set of all non zero real numbers. Then, show that the function \(f:R_0→R_0:f(x)=\frac{1}{x}\) is one-one and onto.

Answer»

To prove: function is one-one and onto

Given: \(f:R_0→R_0:f(x)=\frac{1}{x}\)

We have,

\(f(x)=\frac{1}{x}\)

For, f(x₁) = f(x₂)

\(\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}\)

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

\(f(x)=\frac{1}{x}\)

Let f(x) = y such that \(y\in R_0\)

\(\Rightarrow y=\frac{1}{x}\)

\(\Rightarrow x=\frac{1}{y}\)

Since \(y\in R_0\)

\(\Rightarrow \frac{1}{y}\in R_0\)

⇒ x will also \(\in R_0\), which means that every value of y is associated with some x

∴ f(x) is onto

Hence Proved