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$f: A \rightarrow B$ is onto ~iff~~ $f(A) = B$.
\w

$f: A \rightarrow B$ is onto ~iff~~ $b \in B$ implies there exists an $a \in A$ such that $f(a) = b$.
\w

$f: A \rightarrow B$ is onto ~iff~~ for all $b \in B$, there exists an $a \in A$ such that $f(a) = b$.
\w

$f: A \rightarrow B$ is NOT onto ~iff~~ there exists $b \in B$ s. t. there does not exist 
an $a \in A$ s. t. $f(a) = b$.

\w

Determine if the following functions are onto.  Prove it.  

\w

1.)  $f: R \rightarrow R$, $f(x) = x^2$

2.)  $f: [0, \infty) \rightarrow R$, $f(x) = x^2$

3.)  $f: [0, \infty) \rightarrow [0, \infty)$, $f(x) = x^2$

4.)  $f: R \rightarrow R$, $f(x) = x^3$

5.)  $f: R \rightarrow R$, $f(x) = 2$

6.)  $f: R \rightarrow R$, $f(x) = 8x + 2$

7.)  $f: R \rightarrow R$, $f(x) = x^2 + 3x$

8.)  $f: R \rightarrow R$, $f(x) = e^x$

9.)  $f: R \rightarrow R$, $f(x) = x^4 + x^2$

10.)  $f: R \rightarrow R$, $f(x) = sin(x)$





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For example in N(a/b) = unknot = N(1/0), and $N(z/v) = 6_3 = N(13/8)$

$N\big({13 \over 8}\big) =
 N\big( {-2pq \pm 1 \over -2q^2 }\big)$


Hence (case 1) $13 = -2pq \pm 1$ or (case 2) $-13 = -2pq \pm 1$

Case 1a:  $13 = -2pq + 1$  implies $pq = -6$ ($h = {1 + 1 \over 2} = 1$) 


If $p = 1$,  $q = -6$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {13 \over -72 }\big) \not=N\big( {13 \over 8 }\big)  $

 
If $p = 2$, $q = -3$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {13 \over -18 }\big) =N\big( {13 \over 8 }\big)  $

If $p = 3$, $q = -2$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {13 \over -8 }\big) =N\big( {13 \over 8 }\big)  $


If $p = 6$,  $q = -1$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {13 \over -2 }\big) \not=N\big( {13 \over 8 }\big)  $


Case 1b:  $13 = -2pq - 1$  implies $pq = -7$  ($h = {1 - 1 \over 2} = 0$) 


If $p = 1$, $q = -7$.  $N\big( {-2pq - 1 \over -2q^2 }\big) = N\big( {13 \over -98 }\big) \not=N\big( {13 \over 8 }\big)  $


If $p = 7$, $q = -1$, $N\big( {-2pq - 1 \over -2q^2 }\big) = N\big( {13 \over -2 }\big) \not=N\big( {13 \over 8 }\big)  $



Case 2a:    $-13 = -2pq + 1$ implies $pq = 7$  ($h = {1 + 1 \over 2} = 1$) 


If $p = 1$, $q = -7$.  $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -98 }\big) \not=N\big( {13 \over 8 }\big)  $


If $p = 7$, $q = -1$, $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -2 }\big) \not=N\big( {13 \over 8 }\big)  $


Case 2b:    $-13 = -2pq  - 1$ implies $pq = 6$  ($h = {1 - 1 \over 2} = 0$) 



If $p = 1$,  $q = 6$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -72 }\big) \not=N\big( {13 \over 8 }\big)  $

 
If $p = 2$, $q = 3$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -18 }\big) =N\big( {13 \over 8 }\big)  $


If $p = 3$, $q = 2$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -8 }\big) =N\big( {13 \over 8 }\big)  $


If $p = 6$,  $q = 1$. $N\big( {-2pq + 1 \over -2q^2 }\big) = N\big( {-13 \over -2 }\big) \not=N\big( {13 \over 8 }\big)  $


Hence $(p, q) = (2, \pm 3), (3, \pm 2) $

$U = ({da - jb \over pb - qa} + {j \over p})\circ(h, -1) =  ({d \over -q} + {j \over p})\circ(h, -1) $
  and  $({j \over p} + {da - jb \over pb - qa})\circ(h, -1) = ({j \over p} + {d \over -q})\circ(h, -1) $,

Choosing any $d, j$ such that $pd - qj = 1$:

$(p,q) = (2, 3), (d, j) = (-1, -1)$, $U =  ({-1 \over -3} + {-1 \over 2})\circ(0, -1) = ({1 \over 3} + {-1 \over 2})\circ(0, -1)  $ or 
$ ({-1 \over 2} + {1 \over 3})\circ(0, -1) =   ({1 \over 2} + {-2 \over 3})\circ(0, -1) =  $

$(p,q) = (2, -3), (d, j) = (-1, 1)$, $U =  ({-1 \over 3} + {1 \over 2})\circ(1, -1) $ or $({1 \over 2} + {-1 \over 3})\circ(1, -1) $

$(p,q) = (3, 2), (d, j) = (1, 1)$, $U =  ({1 \over -2} + {1 \over 3})\circ(0, -1) = ({1 \over 2} + {-2 \over 3})\circ(0, -1) $

$(p,q) = (3, -2), (d, j) = (1, -1)$, $U =  ({1 \over 2} + {-1 \over 3})\circ(1, -1) $





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