Define a function $f : \mathbf { R } - \{ 0 \} \longrightarrow \mathbf { R }$ by the formula $f ( x ) = \frac { x + 3 } { x }$ for all nonzero real numbers $x$. Prove that $f$ is one-to-one.

Question

Quizplus · Accepted Answer

The Answer is

f

is one-to-one. Proof: Let

x _ { 1 }

and

x _ { 2 }

be any nonzero real numbers such that

f \left( x _ { 1 } ight) = f \left( x _ { 2 } ight)

. [We must show that

x _ { 1 } = x _ { 2 }

.] By definition of

f

,

\frac { x _ { 1 } + 3 } { x _ { 1 } } = \frac { x _ { 2 } + 3 } { x _ { 2 } } .

Cross-multiplying gives

\left( x _ { 1 } + 3 ight) x _ { 2 } = \left( x _ { 2 } + 3 ight) x _ { 1 } \quad ext { or, equivalently, } \quad x _ { 1 } x _ { 2 } + 3 x _ { 2 } = x _ { 1 } x _ { 2 } + 3 x _ { 1 } .

Subtracting

x _ { 1 } x _ { 2 }

from both sides gives

3 x _ { 1 } = 3 x _ { 2 }

, and dividing both sides by 3 gives

x _ { 1 } = x _ { 2 }

[as was to be shown].

Define a Function f:R−{0}⟶Rf : \mathbf { R } - \{ 0 \} \longrightarrow \mathbf { R }f:R−{0}⟶R

Define a Function $f : \mathbf { R } - \{ 0 \} \longrightarrow \mathbf { R }$