Problem 10 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If the functions \(f(x)\) and \(g(x)\) are defined on \(R \rightarrow R\) such that $$ \begin{aligned} &f(x)= \begin{cases}0, & x \in \text { rational } \\ x, & x \in \text { irrational }\end{cases} \\ &g(x)= \begin{cases}0, & x \in \text { irrational } \\ x, & x \in \text { rational }\end{cases} \end{aligned} $$ then \((f-g)(x)\) is (a) one-one and onto (b) neither one-one nor onto (c) one-one but not onto (d) onto but not one-one

Step-by-Step Solution

Answer

(f-g)(x) is onto but not one-one (d).

Step 1: Understanding Functions

We have two functions. The function \( f(x)\) assigns a value of \(0\) when \(x\) is rational and \(x\) itself when \(x\) is irrational. The function \( g(x)\) assigns a value of \(0\) when \(x\) is irrational and \(x\) itself when \(x\) is rational.

Step 2: Understanding f-g

Now, we need to find \((f-g)(x) = f(x) - g(x)\). We examine this for both rational and irrational \(x\).

Step 3: Case 1: x is Rational

For rational \(x\), \(f(x) = 0\) and \(g(x) = x\). Therefore, \((f-g)(x) = 0 - x = -x\).

Step 4: Case 2: x is Irrational

For irrational \(x\), \(f(x) = x\) and \(g(x) = 0\). Therefore, \((f-g)(x) = x - 0 = x\).

Step 5: Simplifying f-g Function

Thus, the function \((f-g)(x)\) can be expressed as:\[ (f-g)(x) = \begin{cases} -x, & x \text{ rational} \ x, & x \text{ irrational} \end{cases} \]

Step 6: Is f-g One-One?

The function \((f-g)(x)\) is not one-one because there exist distinct values of \(x\) (e.g., a rational \(x\) and an irrational \(x\)) that map to the same value. Consider \(-\sqrt{2}\) when \(x = \sqrt{2}\) (irrational) and \(x = -\sqrt{2}\) (rational); both map to \(\sqrt{2}\).

Step 7: Is f-g Onto?

For any real number \(y\), if \(y\) is rational, there exists an irrational \(x\) such that \((f-g)(x) = x = y\). If \(y\) is irrational, there exists a rational \(x\) such that \((f-g)(x) = -x = y\). Thus, for every \(y\) in \(\mathbb{R}\), there is some \(x\) in \(\mathbb{R}\) such that \((f-g)(x) = y\). Therefore, \((f-g)(x)\) is onto.

Key Concepts

One-One Function

A one-one function, also known as an injective function, is a function where each element of the range is mapped to by exactly one element of the domain. This means that if any two different inputs are mapped to the same output, then those two inputs must actually be the same. In simpler terms, no two distinct inputs result in the same output.

For instance, consider a function where every unique input leads to a unique output. This is the characteristic of a one-one function. However, in the case of the function \((f-g)(x)\) from the exercise, it is not one-one.

• For rational \(x\), the output is \(-x\).
• For irrational \(x\), the output is \(x\).

But you can find distinct inputs—one rational and one irrational—that lead to the same output. Hence, these similarities mean that \((f-g)(x)\) is not one-one.

Onto Function

An onto function, or surjective function, is a type of function where every element of the codomain is mapped to by at least one element of the domain. This ensures that the entire range of possible outputs is utilized. In other words, for each possible output, there is at least one corresponding input.

For the function \((f-g)(x)\), it has been determined that it is onto. What does this imply?

• If \(y\) is a rational number, it can be expressed as \(-x\) for some rational \(x\), fulfilling the condition \((f-g)(x) = y\).
• If \(y\) is an irrational number, there is some irrational \(x\) such that \((f-g)(x) = x = y\).

Thus, every possible real number \(y\) can indeed be obtained from this function, affirming that \((f-g)(x)\) is onto.

Rational Numbers

Rational numbers are numbers that can be expressed as a fraction, where both the numerator and the denominator are integers, and the denominator is not zero. They can be fully written as the quotient of two integers (e.g., \( \frac{3}{4}, 2, -5 \)), and they include both positive and negative integers, fractions, and zero.

When dealing with functions such as \(f(x)\) and \(g(x)\), rational numbers have special assignments. For \(f(x)\), any rational number \(x\) maps to 0. For \(g(x)\), a rational \(x\) maps to itself. This dual nature allows for interesting manipulations in equations that separate rational from irrational values, like creating distinct value ranges for use in functional mapping.

Irrational Numbers

Irrational numbers are real numbers that cannot be expressed as a simple fraction or ratio of two integers. They include numbers such as the square root of a prime number, pi (\(\pi\)), or Euler's number (\(e\)). These numbers have non-terminating and non-repeating decimal expansions.

In the context of functions, irrational numbers play a unique role. For \(f(x)\) in our exercise, when \(x\) is irrational, \(f(x) = x\). Conversely, for \(g(x)\), if \(x\) is irrational, it simply results in 0. This delineation allows us to see how irrational numbers factor into the overall behavior of functions by providing distinct yet predictable outcomes that are separate from rational numbers.