1. ## Simple mapping question

I hope this doesn't belong in the pre-algebra section.

Sorry for so many questions

1. X is the set of elements x, where x is a natural number, and Y = {0,1}. Given that f, the mapping from X to Y, is the remainder when x is divided by 2, determine the following values.

[sol]

f(1)=1
f(2)=0
f(3)=1
f(4)=0
f(5)=1

Can someone explain what this question is asking? I have no idea what this question is asking, like what is "Given that f, the mapping from X to Y, is the remainder when x is divided by 2"???

I also don't get this

2. $f:X \to Y$ is $x \to 2x+1$ --- (1)

In (1), the images of the elements in X differ. (Why? What do they mean by differ anyway?)

How do you know if a function is a one to one mapping? I'm asking how do you figure it out if the two functions share two or more points where they share the same value (parabolas are obvious, but how about cubic functions?)

3. Given that f(n) is the amount of numbers from 1 to n which are relatively prime to n, complete the following exercises.

a) Determine f(2), f(3), f(4), f(5), f(6), and f(7).

I have no idea what the question is asking me. No idea on even how to start.

b) Given that the domain of f(n) is { $n| 2 \le n \ge 7$ }, determine the range.

2. Originally Posted by chengbin
I hope this doesn't belong in the pre-algebra section.

Sorry for so many questions

1. X is the set of elements x, where x is a natural number, and Y = {0,1}. Given that f, the mapping from X to Y, is the remainder when x is divided by 2, determine the following values.

[sol]

f(1)=1
f(2)=0
f(3)=1
f(4)=0
f(5)=1

Can someone explain what this question is asking? I have no idea what this question is asking, like what is "Given that f, the mapping from X to Y, is the remainder when x is divided by 2"???
First of all, a mapping is not something you can make a graph out of per se. It just takes something in (often a number), does something to it, and gives you back something else (often another number...).

The, I suppose "technical", name for this mapping is modulo 2 - if $x$ is even then $x=2y$ for some $y \in \mathbb{N}$ and so $x$ has remainder 0 when divided by 2. Similarly, if $x$ is odd then $x=2y+1$ for some $y \in \mathbb{N} \cup \{0\}$ and so $x$ has remainder 1 when divided by 2.

Let us look at an arbitrary number, $129$. As $129$ is an odd number, f $(129)$ will be $1$ as the remainder when you divide the number by $2$ is just $1$. $129 = 128+1=64*2+1$. Does that make sense?

Now, note that $f:\mathbb{N} \rightarrow \{0,1\}$. This means that the mapping takes numbers from the natural numbers and maps them to the set $\{0,1\}$. Every number is mapped to a single number - this is what a mapping is! Of course, more than one number is mapped to 1, and more than one number is mapped to 0. We call $\mathbb{N}$ - the stuff that we are mappiing from - the domain, and we call $\{0,1\}$ - the stuff we are mapping to - the range (also known as the codomain).

So, a function is injective (in shorthand, 1-1) if $f(x)=f(y)$ implies that $x=y$. That is to say, every element in the range is mapped to by one and only one element. In this example $f(2)=f(4)$ but $2 \neq 4$ and so this mapping is not injective.

Originally Posted by chengbin

I also don't get this

2. $f:X \to Y$ is $x \to 2x+1$ --- (1)

In (1), the images of the elements in X differ. (Why? What do they mean by differ anyway?)
By "differ" they mean that every element in the domain (X) will map to a different element in the range (Y). Basically, $f(x)=f(y) \Rightarrow \ldots \Rightarrow x=y$ (can you prove this? It's not too hard).

Can you see how this relates to the concept of injective (1-1) mappings? I mean, just look at the definition I gave you, above!

Originally Posted by chengbin

How do you know if a function is a one to one mapping? I'm asking how do you figure it out if the two functions share two or more points where they share the same value (parabolas are obvious, but how about cubic functions?)
I'm not entirely sure what you are asking here. Why are you asking how to figure out if two functions share (have?) two or more points where they share the same value? As far as I know, this doesn't really have anything to do with injective (1-1) functions - to see if a function is injective you just have to look at the function itself, not how it relates to other functions...

Originally Posted by chengbin

3. Given that f(n) is the amount of numbers from 1 to n which are relatively prime to n, complete the following exercises.

a) Determine f(2), f(3), f(4), f(5), f(6), and f(7).

I have no idea what the question is asking me. No idea on even how to start.
$x$ is coprime to $y$ if there exists no number other than 1 which divides both $x$ and $y$. For instance, 12 and 5 are coprime while 12 and 9 are not (3 divides both 12 and 9). To find out if two numbers are coprime or not, it is useful to write out their prime decomposition (e.g. $12=2^2.3$, $9=3^3$, etc) and if then the numbers are coprime if and only if they contain none of the same prime divisors. Firstly, note that $f(p)=p-1$ (can you prove why?). So, let us look at a more interesting number - $15$.

(Note that 1 is coprime to every number as no number other than 1 divides 1.)

So, $15=3.5$. Now, $1$ is coprime to $15$, $2$ is coprime, $3$ is not, $4=2.2$ is, $5$ is not, $6=2.3$ is not (note the $3$), $7$ is, $8=2^3$ is, $9=3^2$ is not, $10=2.5$ is not, $11$ is, $12=2^2.3$ is not, $13$ is, and $14=2.7$ is. Counting all the is-es, we see that there are $7$ numbers less than $15$ that are coprime to $15$. Thus, $f(15)=7$.

To find all the numbers quickly, what you need to do is write all the numbers less than your number down - $1,2,3, \ldots, 14$. Then, write down each prime number that divides your number (no repeats though). Then, score off $p$, the first prime in your list, and then 2 $p$, etc, then go to the second prime in your list and do the same, and so on. Then score off 1. All the numbers that are left are the numbers that are coprime! Simples!

So, that is what the mapping does, and the question is asking you to see where $2,3,4,5,6$ and $7$ are mapped to.

Originally Posted by chengbin
b) Given that the domain of f(n) is { $n| 2 \le n \le 7$ }, determine the range.
The domain is the set of numbers that go in, and the range is the set of numbers that come out - it is just all the numbers that are mapped to by $2,3,4,5,6$ and $7$. It will consist of no more that 6 numbers (one for each number in the domain, but getting rid of copies). Basically, the range is $\{f(2), f(3), f(4), f(5), f(6), f(7)\}$ evaluated and with copies of numbers removed (as it is a set - sets do not have multiple copies of the same numbers).

3. Originally Posted by Swlabr
I'm not entirely sure what you are asking here. Why are you asking how to figure out if two functions share (have?) two or more points where they share the same value? As far as I know, this doesn't really have anything to do with injective (1-1) functions - to see if a function is injective you just have to look at the function itself, not how it relates to other functions...
For example, $h:X \to Y$ is $x \to x^3-3x^2-x+3$

This is not a one to one mapping, because this function is not monotone increasing, therefore h(-1)=h(1)=h(3). At these points they share the same point (0). Sometimes 0 is not the only point it shares, because it doesn't reach 0, but reach 1.

Originally Posted by Swlabr
So, $15=3.5$. Now, $1$ is coprime to $15$, $2$ is coprime, $3$ is not, $4=2.2$ is, $5$ is not, $6=2.3$ is not (note the $3$), $7$ is, $8=2^3$ is, $9=3^2$ is not, $10=2.5$ is not, $11$ is, $12=2^2.3$ is not, $13$ is, and $14=2.7$ is. Counting all the is-es, we see that there are $7$ numbers less than $15$ that are coprime to $15$. Thus, $f(15)=7$.
I don't get this explanation for f(15) What is 15=3.5?

Anyway, the solution book says f(2)=1. How can it have 1? The only numbers are 1 and 2, and 1 is a relative prime. 2 doesn't divide into 1.

4. Originally Posted by chengbin
For example, $h:X \to Y$ is $x \to x^3-3x^2-x+3$

This is not a one to one mapping, because this function is not monotone increasing, therefore h(-1)=h(1)=h(3). At these points they share the same point (0). Sometimes 0 is not the only point it shares, because it doesn't reach 0, but reach 1.

I don't get this explanation for f(15) What is 15=3.5?

Anyway, the solution book says f(2)=1. How can it have 1? The only numbers are 1 and 2, and 1 is a relative prime. 2 doesn't divide into 1.
15=3*5 ( a dot can be used instead of a star). So, basically, every number less than 15 where 3 or 5 divide said number is not coprime to 15. These numbers are 3, 6, 9, 12 and 5, 10. There are 6 of them, and 14-6=8 (I think I said 7 in my previous post...which was wrong...).

2 is coprime to 1 as the only number which divides both 2 and 1 is 1. Thus, f(2)=(number of numbers coprime to 2)=1.

I don't understand what the mapping is in question 2. What does the question say precisely?

5. Originally Posted by Swlabr
15=3*5 ( a dot can be used instead of a star). So, basically, every number less than 15 where 3 or 5 divide said number is not coprime to 15. These numbers are 3, 6, 9, 12 and 5, 10. There are 6 of them, and 14-6=8 (I think I said 7 in my previous post...which was wrong...).

2 is coprime to 1 as the only number which divides both 2 and 1 is 1. Thus, f(2)=(number of numbers coprime to 2)=1.
I thought you meant 3.5, or $\frac {7}{2}$ That's why I was confused.

Originally Posted by Swlabr
I don't understand what the mapping is in question 2. What does the question say precisely?
I just had to find if it is a onto mapping or into mapping.

6. Originally Posted by chengbin
I just had to find if it is a onto mapping or into mapping.
Yes, but what mapping?

I have never heard anyone use the word "into" before, although I have used it in my head for the past few years. Can it be used formally, do you know? Like, is it a proper word for injective?

7. Originally Posted by Swlabr
Yes, but what mapping?

I have never heard anyone use the word "into" before, although I have used it in my head for the past few years. Can it be used formally, do you know? Like, is it a proper word for injective?
Here's the definition I got.

Given two sets, X and Y, and the mapping $f: X \to Y$, if each element of Y is an image of at least one element of X, i.e. when f(X)=Y, the mapping is called an onto mapping. Other general mappings are called into mappings.

I'm doing Kumon BTW, and they sometimes does not use the commonly used word for something (like integration by quadrature for Riemann Sums).

8. Originally Posted by chengbin
Here's the definition I got.

Given two sets, X and Y, and the mapping $f: X \to Y$, if each element of Y is an image of at least one element of X, i.e. when f(X)=Y, the mapping is called an onto mapping. Other general mappings are called into mappings.

I'm doing Kumon BTW, and they sometimes does not use the commonly used word for something (like integration by quadrature for Riemann Sums).
What is it you need to do? Do you have a specific mapping you need to find out if it is injective/surjective?