Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define $f: \mathbb{Z}\rightarrow\mathbb{Z}$ by $f(x)= x^2-1$. Then $f$ is a one to one function.

I think it is false because if you put $3$, then $9-1=8$ which you can get it by $2\times 4$ and $1\times 8$.

Could you tell me if I'm wrong or right?

thanks

sincerely

share|improve this question
2  
No, your idea is not right. $2 \times 4$ and $1 \times 8$ have nothing to do with the function $f$. –  Chris Eagle Jun 4 '13 at 21:55
    
I'm curious to know what functions you think are one-to-one by this logic? –  Erick Wong Jun 5 '13 at 2:40
add comment

1 Answer

To prove a function $f$ is not one-to-one, one needs to show that it is NOT the case that for every $x_1, x_2$ in the function's domain,

$f(x_1) = f(x_2) \implies x_1 = x_2$

Put differently, to prove a function $f$ is NOT one-to-one, we need to show there exist $x_1, x_2$ in the function's domain such that $x_1 \neq x_2$, but $f(x_1) =f(x_2)$.


$$f(x) = x^2 - 1$$

$$x_1 = 2: f(2) = 2^2 - 1 = 4 - 1 = 3$$

$$x_2 = -2: f(-2) = (-2)^2 - 1 = 4 - 1 = 3$$

$$f(2) = f(-2) = 3,\;\;\text{but}\;\;x_1 = 2 \neq x_2 = -2$$

Therefore, $f:\mathbb Z \to \mathbb Z$, with $f(x) = x^2 - 1\;$ is NOT a one-to-one function.

share|improve this answer
    
i have another question. a quick question. –  amie Jun 4 '13 at 21:58
    
the set (x E Q l 1<x<2 ) is countable. isnt this false?? –  amie Jun 4 '13 at 21:59
    
No, that's true. One question at a time ;-) –  amWhy Jun 4 '13 at 22:00
    
lol thanks love. but why is it true?? i dont understand –  amie Jun 4 '13 at 22:02
    
amie: that's a separate question. Why don't you post it as a question?: and ask why it's true that it's countable. –  amWhy Jun 4 '13 at 22:05
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.