MHB Prove 1 - 1: Prove Functions are 1 - 1

[JProgrammer]

So I have to either prove that these functions are 1 - 1 or show a counter example to prove they are not. I believe that I have proven that these functions are 1 - 1, but I am not 100% sure:

For each of the following functions, either prove that the function is 1 – 1 or find a counterexample to show that the function is not 1 – 1.
F:R→R
F(x)={(x^2 for x≥0@〖-x〗^2 for x≤0)┤

This function is 1 -1

(x + 1)^2 = x^2 + x +1
-(x – 1)^2 = x^2 – 2x + 1
x^2 + x +1≠ x^2 – 2x + 1

F:Z→Z
F(n)={(n-1 for n even@n^3 for n odd)┤

This function is 1 – 1.

(n + 2) – 1
(n + 2)^3

n^3+6 n^2+12 n+8 ≠ n+1

Is my work sufficient for proving that the functions are 1 - 1? If not, how would I prove that they are 1 - 1? Or am I completely wrong and the functions are not 1 -1?

 

[Evgeny.Makarov]

F:R→R
F(x)={(x^2 for x≥0@〖-x〗^2 for x≤0)

So
\[
f(x)=
\begin{cases}
x^2,&x\ge0\\
(-x)^2,&x\le0
\end{cases}
\]
Are you sure the second line is not $-x^2$ when $x<0$?

This function is 1 -1

(x + 1)^2 = x^2 + x +1
-(x – 1)^2 = x^2 – 2x + 1
x^2 + x +1≠ x^2 – 2x + 1

Why are you considering $(x+1)^2$ and $-(x-1)^2$? What do these expressions have to do with $f$ and the definition of 1-1 function? By the way, $-(x – 1)^2 = x^2 – 2x + 1$ is incorrect.

 

[JProgrammer]

So
\[
f(x)=
\begin{cases}
x^2,&x\ge0\\
(-x)^2,&x\le0
\end{cases}
\]
Are you sure the second line is not $-x^2$ when $x<0$?

Why are you considering $(x+1)^2$ and $-(x-1)^2$? What do these expressions have to do with $f$ and the definition of 1-1 function? By the way, $-(x – 1)^2 = x^2 – 2x + 1$ is incorrect.

I chose x + 1 because x needs to be greater than or equal to 0. I chose x - 1 because x needs to be less than or equal to zero.

 

[Evgeny.Makarov]

I chose x + 1 because x needs to be greater than or equal to 0. I chose x - 1 because x needs to be less than or equal to zero.

Sorry, this makes no sense. First, $x+1$ is not guaranteed to be positive: take $x=-2$ for example. Second, mathematics is all about precision in using definitions and in following the rules of proof. The concept of 1-1 function is not a metaphor; it's a precise statement. Namely, $f$ is 1-1 if for all $x_1$ and $x_2$, the fact that $f(x_1)=f(x_2)$ implies $x_1=x_2$. A proof of statements starting with "For all $z$" usually starts with "Consider an arbitrary $z$". In this case, you consider arbitrary $x_1$ and $x_2$. You are not allowed to assume whether they are positive; they are arbitrary. (The only restriction is that $x_1$, $x_2$ are in the domain of $f$, but here the domain is $\mathbb{R}$, so there is no restriction on $x_1$ and $x_2$.)

After you cut off "For all $x_1$ and $x_2$", the remaining statement is: "the fact that $f(x_1)=f(x_2)$ implies $x_1=x_2$". A proof of "If $A$, then $B$" proceeds as follows: one assumes $A$ and uses it to prove $B$. Here you assume $f(x_1)=f(x_2)$. From there, you need to prove $x_1=x_2$.

Also, please answer the question from post #2:

Are you sure the second line is not $-x^2$ when $x<0$?

This is because the formula for $f$ says $(-x)^2$, but later you consider $-(x-1)^2$. Whether $f$ is 1-1 depends on the answer.

 

[JProgrammer]

Sorry, this makes no sense. First, $x+1$ is not guaranteed to be positive: take $x=-2$ for example. Second, mathematics is all about precision in using definitions and in following the rules of proof. The concept of 1-1 function is not a metaphor; it's a precise statement. Namely, $f$ is 1-1 if for all $x_1$ and $x_2$, the fact that $f(x_1)=f(x_2)$ implies $x_1=x_2$. A proof of statements starting with "For all $z$" usually starts with "Consider an arbitrary $z$". In this case, you consider arbitrary $x_1$ and $x_2$. You are not allowed to assume whether they are positive; they are arbitrary. (The only restriction is that $x_1$, $x_2$ are in the domain of $f$, but here the domain is $\mathbb{R}$, so there is no restriction on $x_1$ and $x_2$.)

After you cut off "For all $x_1$ and $x_2$", the remaining statement is: "the fact that $f(x_1)=f(x_2)$ implies $x_1=x_2$". A proof of "If $A$, then $B$" proceeds as follows: one assumes $A$ and uses it to prove $B$. Here you assume $f(x_1)=f(x_2)$. From there, you need to prove $x_1=x_2$.

Also, please answer the question from post #2:
This is because the formula for $f$ says $(-x)^2$, but later you consider $-(x-1)^2$. Whether $f$ is 1-1 depends on the answer.

Do you mean could x possibly be negative? No it cannot be negative. If x is plugged in as negative, then the other negative will cancel it out. If x is plugged in as a positive, even though there is a negative in front of the x, it will be squared to be a positive.

 

[Evgeny.Makarov]

It's best not to overquote.

Do you mean could x possibly be negative?

No, the question was "Are you sure the second line is not $−x^2$ when $x<0$?". It was not whether $x$ is negative or positive; it's about the formula used in the problem. Is it $-x^2$ or $(-x)^2$ on the second line of the definition of $f$, i.e., when $x<0$? The answer to the problem depends on this. I am asking because your original question uses $(-x)^2$, but in the solution you are using -(x – 1)^2, which supposedly is the result of substituting $x-1$ into $-x^2$ (though this substitution makes no sense in this context).

 

[JProgrammer]

It's best not to overquote.

No, the question was "Are you sure the second line is not $−x^2$ when $x<0$?". It was not whether $x$ is negative or positive; it's about the formula used in the problem. Is it $-x^2$ or $(-x)^2$ on the second line of the definition of $f$, i.e., when $x<0$? The answer to the problem depends on this. I am asking because your original question uses $(-x)^2$, but in the solution you are using -(x – 1)^2, which supposedly is the result of substituting $x-1$ into $-x^2$ (though this substitution makes no sense in this context).

It is (-x^2).

 

[Evgeny.Makarov]

Then the first function is indeed 1-1. You can start proving it as described in post #4.