3. Determine whether each of these functions from is one-to-one/onto. (a) f : R → R, f(x) = 3r. (b) f : Z- Z, f(x) = 5x. (c) f: R Z, f(x) = [r].
3. Determine whether each of these functions from is one-to-one/onto. (a) f : R → R, f(x) = 3r. (b) f : Z- Z, f(x) = 5x. (c) f: R Z, f(x) = [r].
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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
Transcribed Image Text:**Problem 3:** Determine whether each of these functions is one-to-one/onto.
(a) \( f: \mathbb{R} \to \mathbb{R}, \ f(x) = 3x \).
(b) \( f: \mathbb{Z} \to \mathbb{Z}, \ f(x) = 5x \).
(c) \( f: \mathbb{R} \to \mathbb{Z}, \ f(x) = \lfloor x \rfloor \).
**Explanation:**
- **(a)** The function \( f(x) = 3x \) maps real numbers to real numbers. To determine if it is one-to-one, check if different inputs result in different outputs: if \( 3x_1 = 3x_2 \), then \( x_1 = x_2 \) (divide both sides by 3). Thus, it is one-to-one. To check if it is onto, consider that for every real number \( y \), there exists an \( x = \frac{y}{3} \) such that \( f(x) = y \). Therefore, it is onto.
- **(b)** The function \( f(x) = 5x \) maps integers to integers. It is one-to-one because if \( 5x_1 = 5x_2 \), then \( x_1 = x_2 \). However, it is not onto because not every integer is a multiple of 5. For example, there is no integer \( x \) such that \( f(x) = 1 \).
- **(c)** The function \( f(x) = \lfloor x \rfloor \) maps real numbers to integers. This function is not one-to-one, as multiple real numbers can have the same floor value (e.g., \( f(1.2) = 1 \) and \( f(1.9) = 1 \)). It is onto because every integer \( n \) can be represented as \( f(x) \) for \( x \) in the interval \([n, n+1)\).
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