Mathematics - 2016-10-28 (page 1 of 3)

Is anybody familiar with the terms identity relation and reflexive relation? I've understood the difference b/w these two in two different ways and I'd like to know which is correct...

Brody

Hi there @Kaumudi. So how do you think of these terms?

user228700

Hi @Brody :-) My textbook says:

user228700

"A relation $R$ on a set $A$ is said to be reflexive if there exists no element $a$ (belonging to $A$) such that $(a,a)$ doesn't belong to $R$."

user228700

But when I looked this up on the internet, I found this ^ is actually the definition of the identity relation.

user228700

5:15 AM

In addition to this, I also found this:

user228700

"Reflexive = "Every element is related to itself"

user228700

"Identity = "Every element is related to itself only"

user228700

Basically, I'm asking if a reflexive relation is allowed to have elements other than $(x,x)$.

user228700

If it is, then am I correct in my understanding that a reflexive relation can have as many elements as it wants, but it is called a reflexive relation if it contains at least one element of the type $(x,x)$?

user228700

And an identity relation can only have elements of the type $(x,x)$ and every $x$ from the domain $A$ should be included.

user228700

5:21 AM

I hope what I've written makes some sense...

Fargle

@Kaumudi I would say that reflexive means that every possible $(x,x) \in A \times A$ is in $R$. Identity is right though.

user116211

@Kaumudi at least one? it's $\forall x: x\mathcal R x\,.$

user228700

@Fargle Oh, every possible? And there can be other elements too?

Fargle

@Kaumudi Yes--for example, consider the relation $\leq$ on $\Bbb R$. $x \leq x$ always, but $x \leq x+1$ always, too.

user228700

@MAFIA36790 Oh, u missed out the "?". OK...

user116211

5:25 AM

Reflexive relation must contain every $(x,x);$ it doesn't matter if it contains other pairs too.

user228700

@Fargle Hm, OK...

user228700

@MAFIA36790 And the identity relation must contain all ordered pairs $(x,x)$ but it can't contain anything else, yeah?

Fargle

@Kaumudi Correct.

user116211

@Kaumudi yes, damn poor connectivity.

user228700

OK, thanks, guys! :-)

user116211

5:27 AM

@Kaumudi yes.

user228700

@MAFIA36790 B/w ur fingers and the keyboard? :-P

user228700

Ah, I'm just kidding. Anyhoo, thanks!

user116211

I hate when the connectivity is poor.

Brody

If it helps @Kaumudi, "=" is often taught as an example of a reflexive relation.

user228700

Ooh, I have just one more question!

user228700

5:30 AM

@Brody It is also an identity relation so I think it would be very easy to confuse the two, no?

user228700

Is it true that an empty set is a subset of every set but itself?

user228700

..?

Brody

@Kaumudi Sorry was busy

Your intuition that identity is like a "stronger" reflexivity is right. So, "=" being an example of a reflexive relation makes sense.

user228700

OK...

Brody

@Kaumudi Your second question might be too technical (for me), since it involves some pretty foundational stuff.

What do your sources say about $\{\,\}$?

Typically, we say $\{\,\}\subseteq S$ for any set $S$, which includes $\{\,\}$. @Kaumudi

user228700

5:47 AM

It includes $\{\,\}$ as well?

user116211

Can you disprove $\emptyset \not\subseteq \emptyset\,?$

user116211

Can you find some $x\in \emptyset: x\not\in \emptyset\,?$

user228700

@MAFIA36790 Your question is very abstract and it looks like the answer is no, but then again, we're comparing two nothings...

user116211

What I'm saying is that $\emptyset \subseteq \emptyset\,.$

user228700

OK...

user228700

5:50 AM

I have just one last question, dyou mind?

user116211

Empty set is unique, BTW.

user228700

Is this: $y=\sqrt{x}$ a function?

user116211

Let $x= 4;$ what do you get as $y\,?$ Unique value?

Brody

@Kaumudi mmm, that can't be fully answered w/o more info. What sets are involved?

user116211

Hope $\mathcal R: \mathbb R\mapsto \mathbb R\,.$

user228700

5:54 AM

@MAFIA36790 No, well, I mean it's supposed to be like this, right:

user228700

$y=\sqrt{x^2}=|x|$ ?

user228700

@Brody Only $R$.

user116211

@Kaumudi I guess, then yes.

user228700

Then yes, what..?

user228700

That it is a function?

user116211

5:55 AM

it's a function.

Brody

Right. $\sqrt{x}$ typically denotes only the non-negative root. Each $x\in\mathbb{R}$ for which $f$ is mapped to exactly one $y$.

Buuuut...

user228700

OK, how then do we end up with two different roots for an equation like $(x-5)^2=4$?

user228700

So, there are two different roots for that ^, but not for this:

user228700

$\sqrt{(x-5)^2}=4$

user228700

Or is there?

user228700

5:59 AM

I mean, since we dunno what $x$ is, do we remove the | | by taking a ±?

Brody

wait! If you want to strictly follow the definition $$f \text{ is a function iff }\forall x\in X,\exists ! y\in Y,(x,y)\in f\subseteq X\times Y$$ then $f:\mathbb{R}\to\mathbb{R},\; x\mapsto y=\sqrt{x}$ is not a function.

user228700

Huh.

user228700

So do we or do we not generally treat this as a function?

user228700

It seems to me like mostly we don't. Esp. when we've got so many non-linear equations to solve...

user228700

So for all intents and purposes, the square root is not a function, it's a relation, yes?

user228700

6:04 AM

Then why did we define it like this:

user228700

$y=\sqrt{x^2}=|x|$ ?

Brody

It's an issue of technicality at this point. Practically, the principal square root function is indeed a function

user228700

Right, but when dealing with square roots, we aren't talking about this function?

Brody

As written above (with $f:\mathbb{R}\to\mathbb{R}$), for any $x<0$ there is no $y$ at all, which violates the definition

however, if we define $f: [0,\infty)\to\mathbb{R}$ then $f$ is a function

user228700

Sigh. Then what about when solving quadratic equations?!

Brody

6:10 AM

You were on the right track @Kaumudi

user228700

When, before? With the:

user228700

11 mins ago, by Kaumudi

I mean, since we dunno what $x$ is, do we remove the | | by taking a ±?

Brody

$(x-5)^2=4\;\;\Rightarrow\;\;\sqrt{(x-5)^2}=\sqrt{4}\;\;\Rightarrow\;\;|x-5|=2$

user228700

?

Brody

mistakes corrected^

user228700

6:12 AM

OK, this makes sense, but what if I'm trying to solve a quadratic equation and I use the formula $-b±...$?

user228700

In that case, the ± is there 'cause we've taken the square to not be a function, correct?

Brody

yes, it's basically the same thing. to "undo" the absolute value you apply $\pm$, to "undo" a square you apply $\pm\sqrt{}$

Sorry, I'm probably convoluting this @Kaumudi

user228700

What dyou mean to undo?

Brody

apply the inverse to get rid of, e.g. $|x-5|=2\;\;\Rightarrow\;\;x-5=\pm 2$

user228700

Isn't that literally the same as it having two answers? I mean, it switches from being a function to being a relation the minute we place that ± :/

Brody

6:20 AM

You need the $\pm$ or your answer is incomplete ($\to$ wrong). You're right that $\pm\sqrt{}$ is not a function

user228700

OK! So, for all intents and purposes, the square is not a function, yeah?

user228700

The only reason we put that | | in its definition is when defining it as a function, correct?

Brody

If anything $\sqrt{x^2}=|x|$ is a definition for the absolute value function. Just take this next bit in...

The principal square root $\sqrt{}$ operation defines a function, because it only takes the non-negative square root wherever it's defined. The plus-minus square root $\pm\sqrt{}$ operation does not define a function, because it can produce two distinct outputs for the same output.

user228700

OK, and we're mostly using the latter, yeah?

user228700

When solving equations ie.

Brody

6:26 AM

In this context (solving quadratic equations), yes. Because every quadratic has exactly two roots in $\mathbb{C}$. You need the $\pm$ to get them.

user228700

Yes!

user228700

Phew. Thank you! :-D

user228700

Do u have to go right away or is there time for one last (I promise! :-P) question?

Brody

Go ahead :)

user228700

OK, so, for both relations and functions, we need all the elements in the domain to have an image, yeah?

user228700

6:29 AM

It was my understanding that this was the case only w/ functions but I read somewhere just now that indeed, both are defined like this.

user228700

Is this correct?

user116211

@Kaumudi only for the latter.

Brody

@Kaumudi mm Not sure if it's true for relations, so I dunno

user228700

OK, I was thinking about this in terms of a relation given as : Give me a height and I'll give u the name of the person with that height (Excuse all the sketchiness :-P)

user116211

$\mathcal R\subseteq A\times B\,.$

user228700

6:35 AM

So if the domain is $R$, then of course, there will be heights that no person has.

user228700

But in turn, if I define the function as Give me a name and I'll give u a height then yes, every person MUST have an associated height.

user228700

So, I guess that that is a defining characteristic.

user228700

I'm making sense, right?

Brody

might be an awkward example. does every human being have a unique height on continuum? :P

user116211

$F\subset A\times B: \forall x\in B~\exists !~ y\in B: (x, y)\in A\times B\,.$

user228700

6:37 AM

@Brody What dyou mean by on continuum?

user228700

@MAFIA36790 OK, for all.

user228700

OK, I think I'm convinced.

Brody

@Kaumudi Measuring heights on the real number line. Between any two distinct real numbers, there are infinitely many more real numbers.

user228700

@Brody :-P Well...it's convenient, isn't it? Of course, not when we have finite number of people, but oh well, I don't care so much about the details right now.

user116211

@Kaumudi I hate my internet.

user228700

6:40 AM

OK, thank you! @Brody@MAFIA36790!

user116211

There is a typo which I wanted to edit ;/

user116211

It should be $\forall x\in A$ but it is evident.

Brody

@Kaumudi Just being tongue-in-cheek. If we restrict the domain to heights which have a corresponding human being, then the mapping may define a function since it's effectively impossible for two human beings have the same exact "height" over real numbers :P

i.e. the relation is indeed a function (but we hand-waved how "height" is measured or what it precisely means. lot of awkwardness, bleh) @Kaumudi

I do need help with writing vectors as linear combinations... How do I find a nontrivial linear combination for the zero vector?

Rather, say we're working in $\mathbb{R}^n$ and are given $n$ vectors $\mathbf{v}_1,\ldots ,\mathbf{v}_n$ therein. I would guess that $\mathbf{0}=\sum_{i=1}^n c_i \mathbf{v}_i$ only has the trivial solution $c_i=0,\forall i\in[1,n]$. Right?

Brody

7:20 AM

nvm, linear independence is necessary. otherwise, nontrivial solutions may exist

user228700

@Brody Ah, I see what u mean :-)

user228700

@Brody I'm sorry, I graduated high school this year...I'm essentially useless.