Prove that the function $f :\mathbb{Z}\to\mathbb{Z}$ given by $f(x) = 15x − 11$ does not map onto its codomain.

Question

Prove that the function $f : \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = 15x − 11$ does not map onto its codomain.

I am quite confused with this as I'm stuck with how to probably disapprove this statement. I need help with the steps of how I can know when something maps onto its codomain and when it doesn't and why it doesn't.

Answer 1

$15x-11=0$ implies that $15x=11$ and $15$ divides $11$ impossible.

Answer 2

You need to show that fixed some $z \in \mathbb{Z}$ (codomain), the equation $15x - 11 = z$ has no solution for $x \in \mathbb{Z}$. If you find one integer $z$ such that $15x - 11 = z$ has no solution, the result follows.

Answer 3

All you have to do is find an integer $a $ so that $a\ne 15x-11$ for any integer $x $. In other words an example when $a+11$ is not divisible by $15$. So long as $a $ divided by $15$ has any remainder but $4$ the is no $f (x)= a $ so $0,1,2,3,5,6,7,8,etc $ are all numbers that $f $ does $not $ map to.

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Just go to definitions.

A function $f:A\to B$ is onto if for every $b \in B$ there is an $x \in A$ so that $f(x) = b$.

Or in other words, every element of $B$ gets mapped to by some element of $A$. (or as random girl puts it "gets hit"). So... onto: every element gets mapped to. not onto: there are elements that do not get mapped to.

So to prove onto:

You take an arbitrary element $b \in B$ and try to show there is some $x_b \in A$ so that $f(x_b)= b$.

For example of $f:\mathbb R \to \mathbb R$ via $f(x) = x^3$: If for $b\in \mathbb R$ must there be an $x_b$ so the $f(x_b) = x_b^3 = b$? Well, if $x_b^3 = b$ then $x_b = \sqrt[3] b$. Does $x_b = \sqrt[3] b$ exist for all $b$? Yes, yes it does. So $f$ is onto.

But suppose instead we had $g:\mathbb Z \to \mathbb Z$ via $f(x) = x^3$. If $g$ onto. Well, let $b$ be an arbitrary integer. Must there be an $x_b\in \mathbb Z$ so that $f(x_b) = x_b^3 = b$. Again, if $x_b^3 = b$ then $x_b = \sqrt[3]{b}$. Does $x_b = \sqrt[3]{b} \in \mathbb Z$ exist of all integers $b$? No, it does not. It exists only if $b$ is a perfect cube and not all integers are perfect cubes.

Watch out that your domains and codomains are correct though. Consider $h:\mathbb R \to \mathbb Z$ via $h(x) = x^3$. Is $h$ onto? Well, again, for any arbitrary $b \in \mathbb Z$ does there exist a $x_b\in \mathbb R$ so that $f(x_b) = x_b^3 = b$. Well,, again if so that would mean $x_b = \sqrt[3]{b}$. For every integer $b$ does there exist a real $\sqrt[3]{b}$. The answer to that is, yes, yes it does. So $h$ is onto.

To prove something is not onto:

1) It's valid to provide a simple counter example:

Is $g:\mathbb Z \to \mathbb Z$ via $g(x) = x^3$ onto? Well,no if $b= 31 \in \mathbb Z$ and $x^3 = 31$ then $27 < x^3 = 31 < 64$ so $3 < x < 4$ and $x$ is not a perfect square. (That was probably overkill.)

2) If you don't know and aren't sure if there is a counter example, just follow through and see what being onto implies. And see whether that is or is not airtight.

Is $g$ onto? Let $b\in \mathbb Z$ be an arbitrary integer. Let $x_b$ be aninteger so that $x_b^3 = b$ and $x_b = \sqrt[3] b$. Is that true for every $b$? If so it means $b = x_b*x_b*x_b$ must all numbers be in that form. What about $b + 1$. That would mean $b+1 = x_c*x_c*x_c$ and $x_c > x_b$ so $x_c\ge x_b + 1$ so $x^c^3 \ge x^b^3 + 3x_b^2 + 3x_b + 1 > x^b^3 + 1 = b+1$. So both $b$ and $b+1$ can't be mapped to (unless $b=0$ but $b$ was arbitrary so we can assume $b \ne 0$).

.........

So is $f:\mathbb Z\to \mathbb Z$ via $f(x)= 15x -11$ onto.

If $b $ is an arbitrary integer and $x_b$ is such $15x_b -11 = b$ then $x_b = \frac {b+11}{15}=\frac b{15} + \frac {11}{15}$ is an integer. Which it doesn't have to be.

A simple counter example such as $b = 7$ shows it doesn't have to be an integer. Or further more we can show that $\frac {b+11}{15}$ is an integer only if $$b$ has a remainder of $4$ when divided by $15$.