Is the result true if the domain \(\mathbb{R}^{*}\) is replaced by \(\mathbb{N}\) with co-domain remaining \(\mathbb{R}^{*}\)?
Concept / Theory
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To prove that a function is one-one (injective) and onto (surjective), we use the following definitions.
One-One (Injective) Function
A function is one-one if different elements of the domain have different images.
Equivalently, if
implies
then the function is one-one.
Onto (Surjective) Function
A function is onto if every element of the co-domain has at least one pre-image in the domain.
That is, for every
there exists an
such that
Important Observation
The reciprocal function
is defined only when
Hence both the domain and co-domain are taken as
Step-by-step Plan
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Assume two images are equal and prove the corresponding inputs are equal.
Hence establish that the function is one-one.
Take an arbitrary element of the co-domain.
Find its pre-image explicitly.
Hence prove that the function is onto.
Replace the domain by \(\mathbb{N}\) and examine whether every element of \(\mathbb{R}^{*}\) still has a pre-image.
Complete Solution
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- Part I : Prove that the function is One-One
- Let \(x_1,x_2\in\mathbb{R}^{*}\) and suppose \(f(x_1)=f(x_2)\)
- Using the definition of the function,\[\frac{1}{x_1}=\frac{1}{x_2}\]
- Since \(x_1\neq0\) and \(x_2\neq0\) we may multiply both sides by \(x_1x_2\), Therefore,\[x_2=x_1\]
- Hence,\[x_1=x_2\]
- Therefore, whenever\[f(x_1)=f(x_2)\]
- we obtain\[x_1=x_2\]
- Hence, the function is one-one.
- Part II : Prove that the function is Onto
- Let \(y\in\mathbb{R}^{*}\) Since \(y\neq0\)
- consider the element\[x=\frac{1}{y}\]
- Because \(y\neq0\) we have\[x=\frac{1}{y}\neq0\]
- Thus,\[x\in\mathbb{R}^{*}\]
- Now evaluate the function:\[f(x)=\frac{1}{x}\]
- Substitute \(x=\frac{1}{y}\)\[f\left(\frac{1}{y}\right)=\frac{1}{\frac{1}{y}}=y\]
- Thus every element \(y\in\mathbb{R}^{*}\) has a pre-image\[x=\frac{1}{y}\]
- Hence the function is onto.
- Conclusion
- Since the function is both one-one and onto, it is a bijection.
- Part III : When Domain is Replaced by \(\mathbb{N}\)
- Now consider the function \(f:\mathbb{N}\rightarrow\mathbb{R}^{*}\) ddefined by\[f(x)=\frac{1}{x}\]
- Checking One-One
- Suppose \(f(a)=f(b)\) Then\[\frac{1}{a}=\frac{1}{b}\]
- Multiplying both sides by \(ab\)we obtain\[ab\]
- Hence the function is still one-one.
- Checking Onto
- The co-domain is \(\mathbb{R}^{*}\) Take the element \(2\in\mathbb{R}^{*}\)
- If the function were onto, there should exist some \(n\in\mathbb{N}\) such that\[\frac{1}{n}=2\]
- Solving,\[n=\frac{1}{2}\]
- which is not a natural number.
- Hence the element 2 has no pre-image in \(\mathbb{N}\)
- Therefore, the function is not onto
Final Answer
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- The function
\[f:\mathbb{R}^{*}\rightarrow\mathbb{R}^{*},\quad f(x)=\frac{1}{x}\]is one-one and onto
- When the domain is changed to \(\mathbb{N},\)
the function remains one-one but is not onto.
Hence, the original result is not true when the domain is replaced by \(\mathbb{N}\).
Exam Significance
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- This problem is one of the standard applications of the definitions of injective and surjective functions.
- Students learn how to prove one-one and onto properties using formal mathematical arguments.
- The concept of reciprocal functions frequently appears in CBSE board examinations.
- Bijective functions form the foundation for inverse functions, an important topic in higher mathematics.
- Questions of similar pattern are common in JEE Main, NDA, CUET, BITSAT and other entrance examinations.
Key Takeaways
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-
A function is one-one if equal outputs imply equal inputs.
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A function is onto if every element of the co-domain has a pre-image.
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The reciprocal function is well-defined only for non-zero real numbers.
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The function \(f(x)=\frac{1}{x}\) is its own inverse on \(\mathbb{R}^{*}\).
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Changing the domain may preserve injectivity but destroy surjectivity.
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Always examine both the domain and co-domain before deciding whether a function is onto.