1

So I'm trying to prove the function $f(x) = \frac32 x − \frac23$ is bijective (injective and surjective)...

For injection I can easily write $\frac32x - \frac23 = \frac32y - \frac23$; that gives me $x=y$ so it's injective.

So the one more step to prove is to show function is surjective. I can say $f(x)=y$ which gives me $\frac32 x - \frac23 = y$ or $\frac{9x - 4}{6} = y$, which finally gives me $x = \frac{6y+4}{9}$.

But there is a problem when I plug $x$ in function $\frac32 x - \frac23$ I can't get that $y$ value... Basically I get $\frac32 (6y + 4)\frac19 - \frac23 \implies 6y + \frac23 - \frac23$ which is $6y$ but I need $y$? I saw on Wolfram that this function must be surjective but I don't know how to prove it. Can anyone help me with that. And almost forgot to mention $f :\mathbb R \to\mathbb R$.

Thanks in advance :)

Ottavio
  • 2,287
Minotaur666
  • 57
  • 1
    $(3/2)\times 6y/9 = y$. Check your algebra. – Ethan Bolker Dec 10 '20 at 00:46
  • Here is a Mathjax tutorial. Do you mean $\dfrac3{2x}-\dfrac23$ or $\dfrac32x-\dfrac23$? – J. W. Tanner Dec 10 '20 at 00:51
  • 1
    @fleablood to show injective, you show that $f(x)=f(y)$ forces $x=y$. Here, the equation $f(x)=f(y)$ is $\frac{3}{2}x-\frac{2}{3}=\frac{3}{2}y-\frac{2}{3}$. Doing basic algebra reduces the equation to $x=y$. – ndhanson3 Dec 10 '20 at 00:57
  • 1
    "which is 6y but I need y??" That can only happen if you made an arithmetic error. Check your math: $\frac 32\frac {6y+4}9-\frac 23=\frac {18y+12}{18}-\frac 23=\frac {18y + 12- 12}{18} = \frac {18y}{18}=y$. – fleablood Dec 10 '20 at 01:06

1 Answers1

1

The function is surjective, you just made a small computational error.

\begin{align*} \frac{3}{2} \frac{6y+4}{9} - \frac{2}{3} &= \frac{6y+4}{6} - \frac{2}{3}\\ &= y + \frac{2}{3}-\frac{2}{3}\\ &= y. \end{align*}

paul blart math cop
  • 8,066