1. ## [SOLVED] proving bijectivity

Define g: Integers to Integers by g(n)=4n-5. Is g injective? surjective? justify your answer

injectivity
suppose g(n)=g(m) for some m element of integers
$\Rightarrow$ 4n-5=4m-5

$\Rightarrow$ 4n=4m

$\Rightarrow$ n=m
so therefor this statement is injective since each element of g(n goes to an element of g(m)

for surjectivity i know i have to let b E integers and let a E integers since g:integers to integers and i know i have to show that f(a)=b. so:

f(a)=b
4a-5=b
a= $\frac{b+5}{4}$

$f(a)= 4 \frac{b+5}{4} - 5$
$= \frac{4b+20}{4} - 5$
=b+5-5
=b

this is wrong, but i don't understand why. i tried following the books example, but i guess it only works for their specific example.

how am i to show a generic proof for surjectivity in this case? i know that f(a)=2 has no solution in integers (from what my professor wrote on my paper). as well, how do i write the integers symbol using latex?

thank you,

Scott

2. Your mistake is in not realizing that a must be an integer.
If $b=1$ can you find such an a? NO!
Is the function surjective?

3. Originally Posted by scottie.mcdonald
Define g: Integers to Integers by g(n)=4n-5. Is g injective? surjective? justify your answer

injectivity
suppose g(n)=g(m) for some m element of integers
$\Rightarrow$ 4n-5=4m-5

$\Rightarrow$ 4n=4m

$\Rightarrow$ n=m
so therefor this statement is injective since each element of g(n goes to an element of g(m)

for surjectivity i know i have to let b E integers and let a E integers since g:integers to integers and i know i have to show that f(a)=b. so:

f(a)=b
4a-5=b
a= $\frac{b+5}{4}$

$f(a)= 4 \frac{b+5}{4} - 5$
$= \frac{4b+20}{4} - 5$
=b+5-5
=b

this is wrong, but i don't understand why. i tried following the books example, but i guess it only works for their specific example.

how am i to show a generic proof for surjectivity in this case? i know that f(a)=2 has no solution in integers (from what my professor wrote on my paper). as well, how do i write the integers symbol using latex?

thank you,

Scott
The definition for surjectivity is :

FOR all bεZ there exists an aεD(g) ( = domain of g = Z) such that g(a) =b

Now the negation of the above statement is:

There exists a bεZ SUCH that for all aεD(g) g(a)=/=b.

So if you take b=2 you get that,according to the formula g(a) = 4a-5,a=7/4.

That means that if you try ALL the members of your domain none of them will give you the No 2.Only the No 7/4 will give you 2 ,which does not belong to D(g)

Hence the function is not surjective.

Note for surjectivity you must have :

Range of g = Z

4. ahh yes, a must be an integer. *sobs*, i should have noticed that.