# Functions Proof

• Mar 1st 2009, 02:46 PM
1234567
Functions Proof
Hi

Attachment 10373

I know by definition:(lecture notes)

injective, or one-to-one, if different elements of X have different images under F: x1 6= x2 implies F(x1) 6= F(x2) (or equivalently, F(x1) = F(x2) implies x1 = x2).

surjective, or onto, if its range is equal to Y: that is, for every y 2 Y, there is some x 2 X such that F(x) = y.

bijective, or a one-to-one correspondence, if it is both injective and surjective.

But i find it hard to determine wether a function is onto or one-to-one, so can someone please help me.
• Mar 1st 2009, 05:42 PM
GaloisTheory1
Quote:

Originally Posted by 1234567
Hi

Attachment 10373

I know by definition:(lecture notes)

injective, or one-to-one, if different elements of X have different images under F: x1 6= x2 implies F(x1) 6= F(x2) (or equivalently, F(x1) = F(x2) implies x1 = x2).

surjective, or onto, if its range is equal to Y: that is, for every y 2 Y, there is some x 2 X such that F(x) = y.

bijective, or a one-to-one correspondence, if it is both injective and surjective.

But i find it hard to determine wether a function is onto or one-to-one, so can someone please help me.

there is only finitely many to check ie only 7 and 11 to see if 1-1 and onto...so don't be discouraged, this is a quick excercise.

hint part (a), $\displaystyle 2^3=8=1=1^3 \in \mathbb{Z}_7$ and therefore not 1-1.