The formal definitions of the functions

StarShine · 2012-01-11 08:00:30

Hello!

I was on a Maths is Fun website and looked up injective, surjective and bijective. I absolutely understand this as a functional diagram. However, I am stuck on the formal definitions.

Here is what I taken out from the website: "A function f is injective if and only if whenever f(x) = f(y), x = y." - Ok, I got that. I looked at an example of it...

"Example: f(x) = x^2 from the set of real numbers R to R is not an injective function because of this kind of thing:
f(2) = 4 and
f(-2) = 4
This is against the definition f(x) = f(y), x = y, because f(2) = f(-2) but 2 ≠ -2 "

Now, i understand that because the domain containing 2 and -2 both go to 4 in the codomain. On the other hand, what if they were natural numbers? That would be injective and not surjective, am I correct?? Because say you have f(2) = f(3), both go to seperate directions in the codomain, but uncertain if that meets the definition "f(x) = f(y), x = y".

Last edited by StarShine (2012-01-11 08:05:06)

bobbym · 2012-01-11 08:34:01

Hi StarShine;

That would be injective and not surjective, am I correct??

If you are talking about f(x) = x^2 in N then yes it is injective but not surjective. There is no f(x) that yields 3 for instance.

StarShine · 2012-01-11 10:12:46

bobbym wrote:

Hi StarShine;
That would be injective and not surjective, am I correct??
If you are talking about f(x) = x^2 in N then yes it is injective but not surjective. There is no f(x) that yields 3 for instance.

oh right! if it was a rational number, would you just consider whole numbers or think about decimals and fractions too?

Last edited by StarShine (2012-01-11 10:18:05)

bobbym · 2012-01-11 14:41:21

Hi StarShine;

I am not following you there. If it was Q ( rationals ) we were speaking of then that would not be injective or surjective.

Rationals like 4 / 9 would be mapped by 2 rationals ( 2 / 3 ), -( 2 / 3 ) so it is not injective. Some members of the set Q would have no mapping like 17 / 19 for instance. What member of Q could you square to get that?

StarShine · 2012-01-12 02:46:15

bobbym wrote:

Hi StarShine;
I am not following you there. If it was Q ( rationals ) we were speaking of then that would not be injective or surjective.
Rationals like 4 / 9 would be mapped by 2 rationals ( 2 / 3 ), -( 2 / 3 ) so it is not injective. Some members of the set Q would have no mapping like 17 / 19 for instance. What member of Q could you square to get that?

hmm.. this is tricky. I can't think of a member that goes into 17/19 :S Perhaps squaring decimals??? Ughh... fractions are the worst.

bobbym · 2012-01-12 02:53:06

Not all decimals are in Q. Not all decimals are fractions.

is a decimal but it is not in Q.

Ughh... fractions are the worst

Actually, when it comes to computer arithmetic then you can appreciate the value of fractions.

Note; I was editing this while Bob was posting. I left out recurring in my first post. Sorry for that, I do a lot of editing.

Bob · 2012-01-12 02:54:36

or recurring ones.

Bob

StarShine · 2012-01-12 15:40:40

Okay. I think I'm getting there and understanding this . I have one more simple quick question, if a function was real numbers to real numbers (not related to our previous examples above), would you use decimals to prove surjectivity??

bobbym · 2012-01-12 15:53:00

Hi StarShine;

Surjective means every B has at least one matching A ( look at the top diagrams on those pages ). The reals called R have negative numbers too. So when f(n) = n^2 how can you get -1 for instance. You can not. Nothing in R when squared will give you -1 so that function is not surjective.

StarShine · 2012-01-13 14:20:32

bobbym wrote:

Hi StarShine;
Surjective means every B has at least one matching A ( look at the top diagrams on those pages ). The reals called R have negative numbers too. So when f(n) = n^2 how can you get -1 for instance. You can not. Nothing in R when squared will give you -1 so that function is not surjective.

You are right! It would be the same if it was integers. f(n) = n^2 would prove it's injective only, like Real Numbers. Same with positive integers too

bobbym · 2012-01-13 14:26:01

Hi;

With Reals it is not injective either. Because there are 2 inputs that get the outputs.

f(n) = n^2

f(2) = 4 and f(-2) = 4, so it is not injuctive in R ( Reals )

StarShine · 2012-01-15 13:41:01

Ahh... I think I got this. I kinda rushed when I commented back before. Going to do a lot of these practice questions!

bobbym · 2012-01-15 14:12:00

Hi StarShine;

You are doing okay. Brush up on your NZQRC.

StarShine · 2012-01-21 02:10:42

Hi guys, I'm back. I have a quick question. Say, x^2+1 in integer to integer. I was wondering how is this not surjective? Is it because the codomain contains all numbers ...-7,-6,-5,-4....0....1,2,3,4,5..... and 'x^2+1' must hit all numbers in the codomain?

In surjection, does everything from the domain go to every element in the codomain too?

Last edited by StarShine (2012-01-21 02:11:17)

bobbym · 2012-01-21 02:27:08

Hi;

This page explains it all:

http://www.mathsisfun.com/sets/injectiv … ctive.html

Surjective means that every "B" has at least one matching "A" (maybe more than one).

It can not be surjective because no B can be left out. What x for f(x) = x^2 + 1 gives you -5? None!

StarShine · 2012-01-21 02:38:38

Yes you are right. But, what if it was 5x-1 in Real numbers to Real numbers? It's injective and surjective, but how can you get 13 and 2 in this example to prove surjection?

bobbym · 2012-01-21 02:57:20

Hi StarShine;

Just solve for

5x-1 = 13 you will get x = 14 / 5

5x-1 = 2 you will get x = 3 / 5 both are members of R.

StarShine · 2012-01-21 03:48:11

Whaa.. shouldn't 5x be 5*n and then subtract by one?

bobbym · 2012-01-21 03:53:01

Hi;

In post #14 you used 5x-1 so I kept the same variable. It does not matter what name you give the variable, x or n is just the same.

StarShine · 2012-01-21 04:36:28

Ohh sorry. I meant 5*x where x is any number. In post 17, you seem to have divided 14 by 5 to reach 13.

Edit: Does that mean 5*2.8 = 14 -1 = 13

Last edited by StarShine (2012-01-21 04:38:52)

bobbym · 2012-01-21 04:59:38

Hi StarShine;

That is correct.

StarShine · 2012-01-21 05:03:02

Ohhhhh boy. How am I gonna find that out. That will probably take forever without a calculator. Is there a quicker method? Or a link from Maths is fun website?

Last edited by StarShine (2012-01-21 05:03:22)

bobbym · 2012-01-21 05:05:28

Hi StarShine;

You do not need a calculator or an unlimited amount of time. It is easy to solve linear equations of the type

5 x - 1 = 13 by a two step process.

StarShine · 2012-01-21 05:09:03

bobbym wrote:

Hi StarShine;
You do not need a calculator or an unlimited amount of time. It is easy to solve linear equations of the type
5 x - 1 = 13 by a two step process.

Okay I will have a look at linear equations.

bobbym · 2012-01-21 05:20:15

Hi;

Look here and then come back and we will do some more.

http://www.mathsisfun.com/algebra/equat … lving.html

Math Is Fun Forum

#1 2012-01-11 08:00:30

The formal definitions of the functions

#2 2012-01-11 08:34:01

Re: The formal definitions of the functions

#3 2012-01-11 10:12:46

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#4 2012-01-11 14:41:21

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#5 2012-01-12 02:46:15

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#6 2012-01-12 02:53:06

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#7 2012-01-12 02:54:36

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#8 2012-01-12 15:40:40

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#9 2012-01-12 15:53:00

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#10 2012-01-13 14:20:32

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#11 2012-01-13 14:26:01

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#12 2012-01-15 13:41:01

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#14 2012-01-21 02:10:42

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#15 2012-01-21 02:27:08

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#16 2012-01-21 02:38:38

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#17 2012-01-21 02:57:20

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#18 2012-01-21 03:48:11

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#20 2012-01-21 04:36:28

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#21 2012-01-21 04:59:38

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