Question

Show that the function is a one-to-one correspondence. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = mx + b$, $m \ne 0$ Proof. i. Suppose $x, y \in \mathbb{R}$ and $f(x) = f(y)$. Then $mx + b = my + b$, so $mx = my$. Because $m \ne 0$, we have $x = y$. Therefore $f$ is one-to-one. ii. Let $z \in \mathbb{R}$. Choose $t = \frac{z - b}{m}$. Because $m \ne 0$, $t \in \mathbb{R}$. Then $f(t) = m(\frac{z - b}{m}) + b = z$. Therefore, $f$ is onto $\mathbb{R}$.

Name: show that the function is a one to one correspondence f mathbbr to mathbbr given by fx mx b m ne 0 proof i suppose x y in mathbbr and fx fy then mx b my b so mx my because m ne 0 we have x y 89414
Uploaded: 2021-09-03T18:04:34-08:00
Duration: 1 min 5 s
Channel: Carson Merrill
Description: show that the function is a one to one correspondence f mathbbr to mathbbr given by fx mx b m ne 0 proof i suppose x y in mathbbr and fx fy then mx b my b so mx my because m ne 0 we have x y 89414

          Show that the function is a one-to-one correspondence.
$f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = mx + b$, $m \ne 0$
Proof.
i. Suppose $x, y \in \mathbb{R}$ and $f(x) = f(y)$. Then $mx + b = my + b$, so $mx = my$. Because $m \ne 0$, we have $x = y$. Therefore $f$ is one-to-one.
ii. Let $z \in \mathbb{R}$. Choose $t = \frac{z - b}{m}$. Because $m \ne 0$, $t \in \mathbb{R}$. Then $f(t) = m(\frac{z - b}{m}) + b = z$. Therefore, $f$ is onto $\mathbb{R}$.

show that the function is a one to one correspondence f mathbbr to mathbbr given by fx mx b m ne 0 proof i suppose x y in mathbbr and fx fy then mx b my b so mx my because m ne 0 we have x y 89414

Added by Amanda M.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Carson Merrill

09/03/2021

Step 1

A one-to-one correspondence means that the function is both injective (one-to-one) and surjective (onto). Show more…

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Show that the function is a one-to-one correspondence. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = mx + b$, $m \ne 0$ Proof. i. Suppose $x, y \in \mathbb{R}$ and $f(x) = f(y)$. Then $mx + b = my + b$, so $mx = my$. Because $m \ne 0$, we have $x = y$. Therefore $f$ is one-to-one. ii. Let $z \in \mathbb{R}$. Choose $t = \frac{z - b}{m}$. Because $m \ne 0$, $t \in \mathbb{R}$. Then $f(t) = m(\frac{z - b}{m}) + b = z$. Therefore, $f$ is onto $\mathbb{R}$.