How to algebraically prove a function is injective/surjective?

[Qwertyuiop[]]

I know that we can test if a function is injective(one-to-one) using the horizontal line test. If let's say we had a function that we didn't know the graph of or we needed to prove that it's injective or surjective, is there a way to do it algebraically? Like using their definitions? I know a function is injective if for all x, [imath]f\left(x_1\right)=f\left(x_2\right)\:\Rightarrow x_1=x_2[/imath] and a function [imath]f:E \rightarrow F[/imath] is surjective if [imath]\forall y \in F \; \exists x \in E, \; y=f(x)[/imath]. Can you use these definitions or some other method to prove that a function is injective/surjective?

 

[Dr.Peterson]

I know that we can test if a function is injective(one-to-one) using the horizontal line test. If let's say we had a function that we didn't know the graph of or we needed to prove that it's injective or surjective, is there a way to do it algebraically? Like using their definitions? I know a function is injective if for all x, [imath]f\left(x_1\right)=f\left(x_2\right)\:\Rightarrow x_1=x_2[/imath] and a function [imath]f:E \rightarrow F[/imath] is surjective if [imath]\forall y \in F \; \exists x \in E, \; y=f(x)[/imath]. Can you use these definitions or some other method to prove that a function is injective/surjective?

Yes, you can.

Would you like to try an example? Perhaps try proving that [imath]f(x)=\sqrt{x}[/imath] is injective and surjective from the non-negative reals to the non-negative reals, using the definitions you gave.

 

[Steven G]

Why would you think that you couldn't use the definition of injecttive to show that a function is injective?

Why is passing the horizontal test the same as showing that if f(a) = f(b), then a=b?

 

[Steven G]

I'm glad that you liked my comment.
Now I would like to know why is passing the horizontal test the same as showing that if f(a) = f(b), then a=b?

 

[Qwertyuiop[]]

Yes, you can.

Would you like to try an example? Perhaps try proving that [imath]f(x)=\sqrt{x}[/imath] is injective and surjective from the non-negative reals to the non-negative reals, using the definitions you gave.

ok, so for injectivity: for all [imath]x \in \R+[/imath] , [imath]f\left(x\right)=\sqrt{x}[/imath] is injective if [imath]f\left(x_1\right)=f\left(x_2\right)\:\Rightarrow x_1=x_2[/imath]
[imath]\sqrt{x_1}=\sqrt{x_2}[/imath]
[imath]x_1=x_2[/imath] so injective

for surjectivity: If surjective, then [imath]\forall y \in \R+ \: \exists x \in \R+ \; y=\sqrt{x}[/imath]

It's obvious that in R+, the sqrt(x) is surjective as every value of y has a value for x.

 

[Qwertyuiop[]]

I'm glad that you liked my comment.
Now I would like to know why is passing the horizontal test the same as showing that if f(a) = f(b), then a=b?

if it's injective than the line crosses the graph only once, so f(a)=f(b) then a=b.
if it's not injective, the lines cuts the graph more than once, f(a)=f(b) but [imath]a \neq b[/imath].
I don't know if that was the question but that's how i relate the horizontal line test with the injectivity def.

 

[lev888]

if it's injective than the line crosses the graph only once, so f(a)=f(b) then a=b.
if it's not injective, the lines cuts the graph more than once, f(a)=f(b) but [imath]a \neq b[/imath].
I don't know if that was the question but that's how i relate the horizontal line test with the injectivity def.

This needs to be more precise.
Injective: ANY horizontal line that crosses the graph, crosses it in one point.
Not injective: the negation of the above - there exists at least one horizontal line that crosses the graph in 2 distinct points.

 

[Dr.Peterson]

ok, so for injectivity: for all [imath]x \in \R+[/imath] , [imath]f\left(x\right)=\sqrt{x}[/imath] is injective if [imath]f\left(x_1\right)=f\left(x_2\right)\:\Rightarrow x_1=x_2[/imath]
[imath]\sqrt{x_1}=\sqrt{x_2}[/imath]
[imath]x_1=x_2[/imath] so injective

for surjectivity: If surjective, then [imath]\forall y \in \R+ \: \exists x \in \R+ \; y=\sqrt{x}[/imath]

It's obvious that in R+, the sqrt(x) is surjective as every value of y has a value for x.

I think you need to say a little more: there should be either words or symbols stating the connection between lines. In particular, "it is obvious" is not a good statement in a proof.

WHY do you conclude that [imath]x_1=x_2[/imath], and HOW DO YOU KNOW that x exists?

Try again.

 

[Steven G]

if it's injective than the line crosses the graph only once, so f(a)=f(b) then a=b.
if it's not injective, the lines cuts the graph more than once, f(a)=f(b) but a\(\displaystyle \neq\)b.
I don't know if that was the question but that's how i relate the horizontal line test with the injectivity def.

Look at bold above.
So \(\displaystyle f(a)\neq f(a)????\)