Question

i dont understand onto function. plz explain

Answer 1

f:A=(1,3,5),B=(5,7,9)
how f is onto function?

Answer 2

@skullpatrol

Answer 3

@mathmale ur awesome, can you answer this?

Answer 4

onto simply means that all of B is used, or mapped to

Answer 5

\[f=\left(\begin{matrix} 1&3&5\\5&7&9\end{matrix} \right)\]

Answer 6

A maps onto B using the function f

Answer 7

is not it a one one function?

Answer 8

it is 1-1 and onto yes

Answer 9

but not all onto function are 1-1

Answer 10

wt is the difference between one one function nd onto function?

Answer 11

inverses

Answer 12

x^2 is an onto function, but is not 1-1

Answer 13

\[x^2:R\to R^*+\{0\}\]

Answer 14

but \(f(a)\ne f(b)\) for all a,b in R

f(-2) = f(2), but -2 is not 2

Answer 15

isnt R* include 0? the notation eludes me this early lol

Answer 16

give an example which is one one function bt not onto funtion

Answer 17

.. i just did :/

Answer 18

let A={-3,-2,-1,0,1,2,3}
let B={0,1,4,9}
and f:A to B is onto
\[f=\left(\begin{matrix} -3&-2&-1&0&1&2&3\\9&4&1&0&1&4&9\end{matrix} \right)\]

Answer 19

hm

Answer 20

if things are paired 1 to 1, then they make distinct couples. bob is married to sally. when you see sally you know that bob is connected to her. when you see bob, you know sally is connected with him. they are a 1-1 coupling.

but the relationship of the f i gave, x^2 is not 1-1 since we cannot make unique pairings from A to B

Answer 21

-3 is married to 9 ... but 9 is in a relationship with -3 and 3 ...

2 is married to 4, but 4 is in a relationship with -2 and 2

-1 and 1 cannot be determined from 1, only 0 is behaving itself

Answer 22

onto, but not 1-1

Answer 23

the line y=0 is an onto function; it maps the real numbers to 0

but it is not 1-1

Answer 24

is it possible a function is one one but not onto

Answer 25

yes

Answer 26

example?

Answer 27

think of something that uses all the real numbers, but does not give back all the real numbers

Answer 28

hw?

Answer 29

or can you draw a picture of a function that is 1-1, but doesnt use all the elements of the codomian?

Answer 30

no

Answer 31

can e^x produce any negative numbers? or 0?

Answer 32

no

Answer 33

but e^x is in a 1-1 relationship ... its invertible.

but if the codomain is the set of real numbers, then e^x does not produce all the real numbers.

Answer 34

then

Answer 35

\[e^x=\left(\begin{matrix} e^{-3}&e^0&e^2\\e^{-3}&1&e^2&0&-2&-\frac56\end{matrix} \right)\]

Answer 36

e^x is not ONTO the real numbers, but it is 1-1

Answer 37

1-1, but not ONTO