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**atran****Member**- Registered: 2013-07-12
- Posts: 91

Hi,

An **equation** with two variables or more has a **solution set** which is infinite in size.

A **binary relation** between set A and set B is a set of ordered pairs. Set A is the **domain**, and set B is the **codomain**. The relation is a subset of A×B.

A **function** is a relation with the property that each element of its domain is related to exactly one element of its codomain.

Are relation, solution set and **graph** synonyms?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

hi atran

They are not the same but you could certainly find them all in the same piece of work.

Have a look at these pages:

http://www.mathsisfun.com/sets/function.html

http://www.mathsisfun.com/algebra/syste … tions.html

http://www.mathsisfun.com/sets/graph-equation.html

You can get a simple definition of each here:

http://www.mathwords.com/a_to_z.htm

As a relation is between 2 sets you could graph the points, but this would not be possible if the sets were not both numeric

eg (one, 3) (two, 3) (three, 5) ...... (<number as a word>, <number of letters>)

And a solution set might be just a single value, so certainly not the same as the other two.

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**atran****Member**- Registered: 2013-07-12
- Posts: 91

Hi,

Is this thinking correct: A solution set is the set of all n-tuples (including "1-tuple") which are solutions to an equation or a system of equations. A graph is the solution set of a function.

Thanks.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

If an equation has only one unknown and some solutions (n of them) then you could put these solutions in a list and make it an n-tuple. Strictly, that's a single n-tuple, not a set of them.

And if there is more than one unknown then I suppose you could do this:

eg.

a + b + c = 6 where a, b, and c are different counting numbers has this solution set:

A function isn't an equation so it doesn't have a solution set. See

http://en.wikipedia.org/wiki/Graph_of_a_function

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**atran****Member**- Registered: 2013-07-12
- Posts: 91

Thanks for your response.

Regarding functions, I read that all elements of domain should be associated with elements of codomain, which can have more unassociated elements. Is that the same with relations (other than functions), or can a relation have a domain with unassociated elements?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

Yes. To be clear you need to understand the difference between domain, codomain and range. I was asked that here:

http://www.mathisfunforum.com/viewtopic … 49#p237049

And then difference between a relation and a function is covered on many internet sites. This one looks like a simple explanation:

http://www.mathwarehouse.com/algebra/re … nction.php

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**atran****Member**- Registered: 2013-07-12
- Posts: 91

Say I have y=x²-1, where x belongs to the real set, and y to negative integers. Obviously, the graph consists only of one point, which is (0, -1).

Given the restrictions above, is the domain of the relation the set of real numbers or just {0}? Is the codomain equal to the range or the negative integer set?

Are two relations equivalent if they have the same graph but different domains/codomains?

Thanks for help.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

hi atran

To define a function properly one must specify how values are generated (the formula) and give the domain and codomain. You did this with:

Say I have y=x²-1, where x belongs to the real set, and y to negative integers.

So you have answered your own question. The domain is {reals} and the codomain is {negative integers}

As you say there is only one pair of values that exist so the range is {-1}

Are two relations equivalent if they have the same graph but different domains/codomains?

Now that is more tricky to answer. I think it depends on exactly what you mean by 'the same graph'.

I'll try to construct an example. Let's say we have

This is a many : many relationship as every x has two ys and every y has two xs (or none if you pick points beyond 5)

If the domain and codomain are -5 ≤ x , y ≤ +5 and secondly -10 ≤ x , y ≤ +10 then the graphs are the same and the domain/codomain can have different sets.

But what about -2 ≤ x , y ≤ +5 ?

Now only part of the circle is visible in the graph so the same relationship gives different graphs.

It seems to me that relationships must be properly defined in the same way as functions, so that would require the domain and codomain are part of the definition . That would make the answer to your question 'No'. (I think .... )

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**atran****Member**- Registered: 2013-07-12
- Posts: 91

Hi Bob,

I asked you the first question since I was thinking like this:

Say I have y=x² and A={0, 1, -1, √2, -√2, √3, -√3, 2, -2, etc...}. If x∈R and y∈Z then x∈A and y∈N. The first two constraints imply the other two, but is the relation between R and Z or A and N? I'm feeling somehow confused.

Thanks.

*Last edited by atran (2013-09-12 00:32:22)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

hi atran

When you start on a walk, you get to choose whether you put your left foot forward first, or your right. If someone else is setting the rules; (eg. you have joined the army and they want "By the left, quick march!") then you have to play by those rules.

So if you want to invent a function, choose your own domain and codomain. If you are doing an question that has been set, the setter should have told you these sets.

Having said that, there have been a few posts where a question asks the student to say what the domain would be. Although, luckily, it was fairly obvious what was expected, strictly speaking the setter was asking an unreasonable question.

eg. **y = √x**

I could have domain = {x ∈ reals and x ≥ 0} codomain = {reals}

But I could choose domain = {square numbers} codomain = {counting numbers}

So how could someone else know which I was intending unless I say?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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