Your mistake is in not realizing that a must be an integer.
If can you find such an a? NO!
Is the function surjective?
Define g: Integers to Integers by g(n)=4n-5. Is g injective? surjective? justify your answer
suppose g(n)=g(m) for some m element of integers
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:
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?
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