# Proving a function is onto.

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• Oct 11th 2011, 04:10 PM
mathproblems
Proving a function is onto.
Hello Everyone,
I need help with y =2^3-4. To prove it it 1-to 1, onto or both.
I think I understand how to to prove that the function is one to one.
2^3-4=2^3-4 then +4 and then divide by 2, so we have x(1)^3=x(2)^3. They are even, if and only if x(1)= x(2) are the same. Therefore this is One to One function.

What about Onto function y =2^3-4.
Do I prove it by counterexample? Or otherwise? I can see that it's onto ....but how to write a proof.
Please someone help me! Thank you.
• Oct 11th 2011, 05:01 PM
vandrop
Re: Please someone help me! with ONTO function
y=2^3-4=5

(Are you sure you're typing this right?) There is only one value of y. Thus if one value of x satisfies the equation, the equation is onto... however, in this case, every value of y is 5 so it wouldn't be one-to-one.

If you mean y=x^3-4, this is one-to-one. You just have to prove y = n for some x. (HINT HINT Plug in for y).
• Oct 11th 2011, 09:41 PM
mathproblems
Re: Please someone help me! with ONTO function
Correction. Sorry. Forgot n ^3

______________________
Domain and codomain all real number.

I need help with f(n) =2n^3-4. To prove it it 1-to 1, onto or both.
I think I understand how to to prove that the function is one to one.
2n^3-4=2n^3-4 then +4 and then divide by 2, so we have n(1)^3=n(2)^3. They are even, if and only if n(1)= n(2) are the same. Therefore this is One to One function.

What about Onto function f(n) =2n^3-4?
Do I prove it by counterexample? Or otherwise? I can see that it's onto ....but how to write a proof.

Thank you.
• Oct 12th 2011, 02:31 AM
Plato
Re: Proving a function is onto.
Quote:

Originally Posted by mathproblems
[FONT=Comic Sans MS]I need help with $\displaystyle y=f(x) =2x^3-4$. To prove it it 1-to 1, onto or both.

Please do not use special fonts.
If $\displaystyle f(a)=f(b)$ then $\displaystyle 2a^3-4=2b^3-4$.
If that means that we must have $\displaystyle a=b$ then it is one to one.

If $\displaystyle c\in \mathbb{R}$ what is $\displaystyle f\left( {\sqrt[3]{{\frac{{c + 4}}{2}}}} \right)~?$
Is it onto.
• Oct 12th 2011, 07:24 AM
mathproblems
Re: Proving a function is onto.
Sorry, I dont get it. How is it onto? And why c+4?
• Oct 12th 2011, 07:31 AM
Plato
Re: Proving a function is onto.
Quote:

Originally Posted by mathproblems
Sorry, I dont get it. How is it onto? And why c+4?

Just evaluate $\displaystyle f\left( {\sqrt[3]{{\frac{{c + 4}}{2}}}} \right)$.

What do you get?
• Oct 12th 2011, 07:39 AM
mathproblems
Re: Proving a function is onto.
I don't get how you got c+4 and why all of it inside 3 roat. would you please write down transition before f(). Thank you.

If I evaluate, I plug 1 or -1, result is a real number. Then it's onto.
• Oct 12th 2011, 07:39 AM
TheChaz
Re: Proving a function is onto.
Quote:

Originally Posted by Plato
Please do not use special fonts....

What's the problem? Soroban does it - I would be shocked if he didn't in any future post...
• Oct 12th 2011, 07:45 AM
Plato
Re: Proving a function is onto.
Quote:

Originally Posted by TheChaz
What's the problem? Soroban does it - I would be shocked if he didn't in any future post...

I do not reply to or edit his posts. I find is distracting when replying.
• Oct 12th 2011, 07:48 AM
Deveno
Re: Proving a function is onto.
what Plato is getting at is this:

the way you show a function is onto, is show if you pick any old point in the co-domain, you can find some point in the domain that maps to it.

since the co-domain is the real numbers, if we want to prove it is onto, we must find, for every real number c,

a real number x with f(x) = c. Plato has given you a BIG hint on what this x might be.
• Oct 12th 2011, 07:53 AM
Plato
Re: Proving a function is onto.
Quote:

Originally Posted by mathproblems
Sorry, I dont get it. How is it onto? And why c+4?

Do you not see that $\displaystyle f\left( {\sqrt[3]{{\frac{{c + 4}}{2}}}} \right)=c~?$.
Never mind where it comes from.
That proves that given any $\displaystyle c\in\mathbb{R}$ there is a real number that is mapped to c.
So the function is onto.

Now for from it came.
If $\displaystyle c=2x^3-4$ can you solve for $\displaystyle x~?$
• Oct 12th 2011, 08:00 AM
TheChaz
Re: Proving a function is onto.
Quote:

Originally Posted by Plato
I do not reply to or edit his posts. I find is distracting when replying.

Indeed, once I tried to parse a quote of his in my reply... I learned that my LaTex skills are far inferior!
• Oct 12th 2011, 08:03 AM
mathproblems
Re: Proving a function is onto.
I am trying to get better at at...I really want to understand how f of 3 roat appeared...;-(
• Oct 12th 2011, 08:15 AM
Plato
Re: Proving a function is onto.
Quote:

Originally Posted by mathproblems
I am trying to get better at at...I really want to understand how f of 3 roat appeared...;-(

Are you saying that you cannot solve $\displaystyle c=2x^3-4$ for $\displaystyle x~?$
If so, there is no wonder you are having difficulty with This.
• Oct 12th 2011, 08:40 AM
mathproblems
Re: Proving a function is onto.
yes, this is what I am saying. I went to school 25 years ago...now I don't remember some things. And now I am taking discrete math ...
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