Show that [4,5] and [4,12] are equivalent or have the same cardinality

Question

I have 2 intervals, $[4,5]$ and $[4,12]$, so far I've just have found the bijection between the two which is $f: [4,5] \to [4,12]$, $f(x) = 8x-28$.

If you can, make a full example proof step by step. Edit : Thanks to quasi and fleablood for providing a neat proof

Answer 1

Let $f\:[4,5] \to [4,12]$ be defined by $f(x) = 8x - 28$.

Edit:

In light of fleablood's answer, the statement

$\qquad$Let $f:[4,5] \to [4,12]$ be defined by . . . .

needs some verification.

Specifically, we need to show that for any input $x$ in $[4,5]$, the output $f(x)$ is in $[4,12]$.

Thus, for $x \in [4,5]$, let $f(x) = 8x-28$. \begin{align*} \text{Then}\;\;&x \in [4,5]\\[4pt] \implies\;&4 \le x \le 5\\[4pt] \implies\;&32 \le 8x \le 40\\[4pt] \implies\;&4 \le 8x-28 \le 12\\[4pt] \implies\;&4 \le f(x) \le 12\\[4pt] \implies\;&f(x) \in [4,12]\\[4pt] \end{align*} Therefore $f$ is a function from $[4,5]$ to $[4,12],\;$i.e., $f:[4,5] \to [4,12]$.

end Edit

The goal is to show $f$ is a bijection.

First, we show $f$ is injective.

Suppose $f(x) = f(y)$, for some $x,y \in [4,5]$.

\begin{align*} \text{Then}\;\;&f(x) = f(y)\\[4pt] \implies\;&8x-28 = 8y-28\\[4pt] \implies\;&8x = 8y\\[4pt] \implies\;&x = y\\[8pt] & \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Therefore $f$ is injective.} \end{align*}

Next, we show $f$ is surjective.

Let $b \in [4,12]$, and let $a = {\large{\frac{b+28}{8}}}$.

Claim $a \in [4,5]$, and $f(a) = b$. \begin{align*} & \!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{First we show $a \in [4,5]\,$:}\\[4pt] &b \in [4,12]\\[4pt] \implies\;&4 \le b \le 12\\[4pt] \implies\;&32 \le b+28 \le 40\\[4pt] \implies\;&4 \le \frac{b+28}{8} \le 5\\[4pt] \implies\;&4 \le a \le 5\\[4pt] \implies\;&a \in [4,5]\\[4pt] &\text{as claimed.}\\[12pt] & \!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Next we show $f(a)=b\,$:}\\[4pt] f(a) &= 8a - 28\\[4pt] &=8\left(\frac{b+28}{8}\right)-28\\[4pt] &=(b+28)-28\\[4pt] &=b\\[4pt] &\text{as claimed.}\\[12pt] & \!\!\!\!\!\!\!\!\!\!\!\!\!\! \text{Therefore $f$ is surjective.}\\[4pt] \end{align*} Since $f$ is both injective and surjective, $f$ is bijective.

Remark: We found $a$ by solving the equation $f(a)=b$ for $a$, but that was side work. The proof doesn't need to show the details of how we found $a$. It only needs to show that the $a$ we chose actually works. Of course, if someone asks "How did you figure out what expression to use for $a$?", you can show the side work.

Answer 2

HINT. 1) Draw an angle between $0^{\circ}$ and $90^{\circ}$.

2) enter the intervals $[a,b]$ and $[A,B]$ in such a way that their ends touch the sides of the angle.

3) a line that passes through the vertex of the angle in its interior cuts $[a,b]$ in $x$ and $[A,B]$ in $f(x)$.

4) The geometric intuition tells you that $f$ is a bijection and the analytic form of this function depends on the position you give to the figures in the coordinate plane.

Answer 3

To prove $f(x) = 8x -28$ is a bijection you need to prove three very easy things.

1) $f: [4,5]\to [4,12]$. Or in other words: $x \in [4,5]\implies f(x) \in [4,12]$

2) $f$ is injective. Or in other words $f(x) = f(y) \implies x = y$.

3) $f$ is surjective. Or in other words: if $y \in [4,12]$ then is an $x\in [4,5]$ so that $f(x) = y$.

===

Proof 1) If $4 \le x \le 5$ then

$32 \le 8x \le 40$ and

$ 4 \le 8x -28 \le 12$ and so $4 \le f(x) \le 12$.

$4= 8*4 - 28 \le 8x -28=f(x) \le 8*5 - 28 = 12$.

So $x \in [4,5] \implies f(x) \in [4,12]$.

2) If $f(x) = f(y)$ then $8x - 28 = 8y - 28$ and therefore $x =y$.

3) For any $y$ so that $4 \le y \le 12$, if $f(x) = 8x-28 = y$ then $x =\frac {y+28}8$.

And if $4 \le y = 8x -28 \le 12$ then

$32 \le 8x \le 12$

$4 \le x \le 5$.

For every $y \in [4,12]$ there is an $x = \frac {y + 28}8 \in [4,5]$ so that $f(x) = y$.