Why is sin(3pi/2) not equal to 1?

[nacho-man]

What exactly does it mean for something to be closed under addition.

most of the examples and questions i have gone past, are answered with

u + v = u1+v1, ... , u_n + v_n and hence this is closed under addition.

So i tried looking for examples where something was not closed under addition, and I found this [refer to attached image]

I follow everything up until the second last and last line;

"For I have sin(3pi/2) = - 1 $\neq$ 1

what does it matter that sin(3pi/2) doesn't = 1?

Would anyone be able to explain what is going over my head? It seems like such a basic concep that I am not understanding.

View attachment 2410

 

[I like Serena]

Hi nacho!

Let me give it a shot to try and clarify.

We want to know if $(3\pi/2, 1)$ is an element of W.

For it to be the case, it must be of the form $(u, \sin u)$.
So $u = 3\pi/2$ and $\sin u = 1$.

If we substitute the first in the second, we get:
$$\sin 3\pi/2 = 1$$
But this is not true.
Therefore $(3\pi/2, 1)$ is not an element of W.

So we have found a counter example where the sum of 2 elements in W is not an element of W.

 

[Prove It]

For a set to be "closed" under an operation (such as addition or multiplication), it means that whenever you add two numbers in that set, you will always get another number that belongs to that set.

For example, the set of all real numbers is closed under addition, because when you add any two real numbers you always get a real number.

As another example, the set of all odd integers is NOT closed under addition, because when you add two odd numbers you get an EVEN number, something not in that set.

 

[Deveno]

Here is an often-used example:

Suppose $S \subseteq \Bbb R^2 = \{(x,0): x\in \Bbb R\} \cup \{(0,y): y \in \Bbb R\}$

(that is to say the $x$-axis and the $y$-axis together).

Now, it is clear that two points on the $x$-axis added together still lie on the $x$-axis:

$(x_1,0) + (x_2,0) = (x_1+x_2,0)$, and similarly for the $y$-axis.

But if we add a point on the $x$-axis, to a point on the $y$-axis, say for example:

(2,0) + (0,3) = (2,3)

it is clear that this point does not lie on EITHER axis (unless both points are the origin, a special case).

In general, we are often given a SUBSET of a vector space, and we want to know if it includes all possible (finite) linear combinations of elements in the set. Put another way, we want to know if the addition operation of some vector space $V$, "stays within $S$", so that addition is an operation on $S$ in its own right. And this means:

IF $u \in S$, AND $v \in S$, THEN $u+v \in S$ also.

Subsets that do not obey this are said to be "not closed under addition".

 

[nacho-man]

Hi thanks for the responses everyone, I seem to have had a similar conceptual understanding, but would like to see it applied in a more general sense.

I have attached an example of what I mean.

How is it conclusive from the third and fourth line that the set W is closed under addition and multiplication?

I mean, all they've done is added them up and multiplied by scalar k. But can't you do that with any general set, and then turn around and say it is closed under scalar multiplication and addition?

 

[Deveno]

Hi thanks for the responses everyone, I seem to have had a similar conceptual understanding, but would like to see it applied in a more general sense.

I have attached an example of what I mean.

How is it conclusive from the third and fourth line that the set W is closed under addition and multiplication?

I mean, all they've done is added them up and multiplied by scalar k. But can't you do that with any general set, and then turn around and say it is closed under scalar multiplication and addition?

Well, no, with some sets that doesn't work.

Here is a more sophisticated example:

We can form a vector space out of the set of real polynomial functions of degree less than or equal to two (this space is often called $P_2$ but it has dimension THREE), by defining, for:

$p_1(x) = a_0 + a_1x + a_2x^2$ and $p_2(x) = b_0 + b_1x + b_2x^2$ (here, $a_0a_1,a_2,b_0,b_1,b_2$ are all real numbers (constants)).

that:

$(p_1+p_2)(x) = (a_0 + b_0) + (a_1+b_1)x + (a_2+b_2)x^2$

$(cp_1)(x) = ca_0 + (ca_1)x + (ca_2)x^2$.

This has dimension 3, because one possible basis is $\{1,x,x^2\}$.

Now suppose we take the subset $S = \{p \in P_2: p(1) = 1\}$. We will see $S$ is not closed under addition. To do so, it suffices to find $f,g \in P_2$ with $f(1) = g(1) = 1$, but $(f+g)(1) \neq 1$.

Well one obvious such polynomial is the constant polynomial $f(x) = 1$, for all real $x$. Another is the linear polynomial $g(x) = x$.

Now $(f+g)(x) = 1 + x$, and we see that $(f+g)(1) = 1 + 1 = 2 \neq 1$.

So $S$ is not closed under addition, and as a result $S$ is not "big enough" to be a subspace.

HOWEVER, for ANY set $S$, we can always take $\text{span}(S)$ which includes ALL linear combinations (over our given field, usually the real numbers, but not always), and such a "span-set" will ALWAYS form a subspace.

**********

Let's look at a very simple vector space, the Euclidean plane $\Bbb R^2$ (if you understand this vector space well, you have a good platform to generalize from). Let's suppose we have a subset $S$ of the plane, that satisfies 3 properties:

1. Closure under addition:

if $(x,y),(x',y') \in S$ so is $(x+x',y+y')$

2. Closure under multiplication:

if $(x,y) \in S$ and $c$ is any real number, then $(cx,cy) \in S$.

3. $S$ has at least one vector (point) in it.

What would such a set look like?

Well two possibilities immediately present themselves:

a) The set consisting of just $\{(0,0)\}$. This is called a TRIVIAL subspace.
b) All of $\Bbb R^2$ itself (duh).

Let's suppose that we have two elements $(a,b),(a',b') \neq (0,0) \in S$ such that $(a',b') \neq (ca,cb)$ for any $c$.

I leave to you (or you can take it on faith)to show that in such a case $ab' - a'b \neq 0$ (hint: it's easier to show that if $ab' - a'b = 0$, that: $b' = cb, a' = ca$ for some $c$. Consider the cases $a = 0$ and $a \neq 0$ separately).

Now by (2), both $a'(a,b)$ and $(-a)(a',b')$ are in $S$, thus by (1), so is their sum:

$a'(a,b) + (-a)(a',b') = (aa',ba') - (aa',ab') = (0,a'b-ab')$.

Since $a'b - ab' \neq 0$ if we call this number $d$, we have:

$\frac{1}{d}(0,d) = (0,1) \in S$, by (2).

A similar argument (which you should provide) shows $(1,0) \in S$.

Finally, we see that any point $(x,y) = x(1,0) + y(0,1) \in S$, by taking (1) and (2) together.

So in this case, we see that two such points force $S$ to be all of $\Bbb R^2$, if $S$ is to satisfy our two closure conditions.

Are there any OTHER such sets $S$?

Let's try to find a minimal one "bigger than just the origin".

So we have at least one point (which is not the origin), say $(a,b) \in S$.

By (2) we have to have all points $(ca,cb) \in S$ for any real number $c$. Is this big enough? That is, we suppose:

$S = \{(ca,cb): c \in \Bbb R\}$, and since (2) is satisfied automatically, we just need to see if (1) is.

So, we pick two points in $S$, say:

$(x_1,y_1) = (c_1a,c_1b)$ and
$(x_2,y_2) = (c_2a,c_2b)$, and we see if $(x_1,y_1) + (x_2,y_2) \in S$.

Now $(x_1,y_1) + (x_2,y_2) = (c_1a,c_1b) + (c_2a,c_2b) = c_1(a,b) + c_2(a,b) = (c_1+c_2)(a,b) = ((c_1+c_2)a,(c_1+c_2)b)$.

Since $c_1+c_2$ is a real number if $c_1,c_2$ are, this shows the sum is indeed in $S$. So we do have closure (of addition).

What does such a set look like?

Since $(a,b) \neq (0,0)$ one of $a,b \neq 0$. Suppose that $a \neq 0$ (I leave it to you to puzzle out what happens if $a = 0$).

In this case, we have, for any point $(x,y) = (ca,cb) \in S$, that:

$y = \frac{b}{a}x$, that is, $S$ is the line through the origin with slope $\frac{b}{a}$.

What does this mean?

It means that any subset of the Euclidean plane closed under addition and scalar multiplication that has at least one point in it is either:

1. Just the origin
2. A line
3. The whole plane

This is why "linear algebra" is so-named. The "linearity" subspaces have is due to the "line-ness" (or in higher dimensions, the "plane-ness") of the subspaces.

 

[I like Serena]

How is it conclusive from the third and fourth line that the set W is closed under addition and multiplication?

I mean, all they've done is added them up and multiplied by scalar k. But can't you do that with any general set, and then turn around and say it is closed under scalar multiplication and addition?

First off, your example contains a mistake.
As the first step a definition is "repeated" for W.
But that is not the W from the problem statement, nor are the remaining steps applicable to it.
Confusing! (Doh)
Anyway, the point is that for any 2 elements of W the addition is an element of W.
So we pick 2 element with symbols that can still represent any element of W.
Then we add them and verify that the result matches the definition of an element in W.
It must then be true that the addition of any 2 elements from W yield indeed an element in W.

Similarly we multiply a representation of any element of W, with a scalar that can be any scalar. And again we verify that the result matches the definition of an element in W.

That makes it closed for addition and scalar multiplication.

 

[nacho-man]

First off, your example contains a mistake.
As the first step a definition is "repeated" for W.
But that is not the W from the problem statement, nor are the remaining steps applicable to it.

Not exactly sure what you mean here?

Anyway, the point is that for any 2 elements of W the addition is an element of W.
So we pick 2 element with symbols that can still represent any element of W.
Then we add them and verify that the result matches the definition of an element in W.
It must then be true that the addition of any 2 elements from W yield indeed an element in W.

This seems consistent with what my friend was saying.
So does this mean to say, for example let our
subset be in the form:

W =
\left[ {\begin{array}{cc}
a & b \\
c & d \\
\end{array} } \right]

Then, if hypothetically, after adding two subsets of W, we get
M=
\left[ {\begin{array}{cc}
a+b & b \\
c & d \\
\end{array} } \right]
Then W is not closed under addition, since it is in a different form?
WhereasI know the second matrix I specified isn't exactly true, but if I were to take 2 matrices in the same form as our original subset, add them together and then get a result something in which the format of the entries does not line up with the format of the original entries, that means it's not closed under addition?


I don't think I get it yet.
I understand the theory behind it and what it means, but I fail to see how one can just take arbitrary values, add them and then say the resulting is closed under addition.

Could anyone provide a general example with arbitrary values where this would not work? Or would the subset have some specific characteristics

 

[Deveno]

Sure, let:

$W = \left\{\begin{bmatrix}a&b\\c&d \end{bmatrix} \in \text{Mat}_{2 \times 2}(\Bbb R): b \text{ or } c = 0\right\}$

Prove this is NOT closed under addition.

 

[I like Serena]

Not exactly sure what you mean here?

Look again at your problem statement and how its resolution starts? (Thinking)

 

[nacho-man]

Look again at your problem statement and how its resolution starts? (Thinking)

Ah I see what you mean, just glossed over that.

Also thanks deveno, somehow I must have not noticed your previous post but that makes sense now! :)

Thanks everyone I really appreciate the help I get on this forums.