Question

1

Answer 1

do you understand what it is you need to do ?

Answer 2

i will take that as a no
you have to solve
\[x\oplus y = x \] for \(y\)

Answer 3

yep, y will be -1 someone helped me. I don't understand second part

Answer 4

@satellite73 can you check here: http://openstudy.com/users/best_mathematician#/updates/5404a1e0e4b0f2ed1e14518a

Answer 5

what is the second part?

Answer 6

what does -x mean in this context

Answer 7

oh in this case the "zero" element, i.e. the additive identity is \(-1\) i guess

Answer 8

\[x+y+1=x\iff y=-1\] so that is the additive identity
to find the additive inverse, solve
\[x\oplus y=-1\] for \(y\)

Answer 9

-2 -x

Answer 10

just like if it was regular old 0 you would solve
\[x+y=0\] and get \(y=-x\)

Answer 11

maybe , lets try it
\[x+y+1=-1\\
x+y=-2\\
y=-2-x\] ok looks good

Answer 12

still it didn't answer my question. what does -x represent? zero element? how? I don't want answers I really wanna understand what's going on

Answer 13

it is written in english, not math, so maybe it is ambiguous
in regular old numbers \(0\) is the additive identity because \(a+0=0+a=a\)
in this system where \(x\oplus y=x+y+1\) the additive identity is \(-1\) because
\[-1\oplus x = x\oplus -1=x\]

Answer 14

in regular old numbers the additive inverse of \(x\) is denoted by \(-x\) because \(x+(-x)=0\) the identity

Answer 15

in this system the additive inverse of \(x\) is \(-2-x\) because
\[x\oplus -2-x=-1\]

Answer 16

so taking the notation of sticking a minus sign in front of a number to make it the additive inverse, you see that since the additive inverse of \(x\) is \(-2-x\) you use the same notation and call it \(-x\)
it is a bit confusing

Answer 17

should probably call it something else, since you have a \(-x\) twice, once in \(-2-x\). they mean different things

Answer 18

what if we replace -x in the equation we will get is x - x + 1 = 1, so dont knot but -x should mean one of those 8 property in these context

Answer 19

that is why it is confusing
one symbol \(-x\) should not mean two different things
i would just say it in engish
the additive inverse of \(x\) is \(-2-x\) and leave it at that

Answer 20

i must say one more thing, there is a question above that which asks the same question what does -x mean. and the answer is it means 1/x

Answer 21

Just a poorly written question.

Answer 22

If you use -x as it should be (not as a meaning for additive inverse), I don't know what it signifies , or what special significance it has as I mentioned before :(

Answer 23

also the question asks to show S to be a vector space but I ignored that part, if that helps

Answer 24

It doesn't appear to be a vector space though if you check \(\large 1\otimes x =^? ~ x\)
\(1 \otimes x = 1x+1 = x+1\ne x\)

Answer 25

i am so confused....hahaa

Answer 26

which part are you confused by?

Answer 27

I am not sure what kirby is saying but if they say its a vector space im sure it is. you need to show that all the axioms hold.

Answer 28

what is this

c(x) X = cx + c.

Answer 29

oh the multiplication ... got it

Answer 30

idk it is second operation i guess

Answer 31

I was trying to say that one of the properties of a vector space is that \(1\vec{x}=\vec{x}\) which doesn't seem to hold here.

Answer 32

right, i didnt notice this.

Answer 33

well, its not that 1*x = x , its that there exists an element we will call it 1, such that 1*x = x

Answer 34

but it need not be actually \(1\in \mathbb{R}\)

Answer 35

but 0 just means additive identity...

Answer 36

so the multiplicative identity is cx+c = x iff c(x+1)=x iff c=x/(x+1)

Answer 37

so here we have \("1" = \frac{x}{x+1 }\)

Answer 38

which makes -x that?

Answer 39

dont think of it as -x. It is asking for the element that you need to add to x in order to get the additive identity.

Answer 40

gotcha

Answer 41

i.e.

x+ ? = -1

so
x+?+1 = -1
solving for ? we get
? = -2-x as @satellite73 suggested

Answer 42

that first + should have a circle around it...

Answer 43

now you need to find the multiplicative inverse, multiplicative identity, and check commutativity, and associativity and the other axioms. let me know if you need help on one of those...

Answer 44

-x is often used to mean, in general, the additive inverse...
and x^(-1) means multiplicative inverse. Often they will not be the same element as if there were being computed in R. So it makes it very tricky to say something like -x = -2-x
because on the right hand side we mean the additive in verse of x in R, but on the left side we mean the additive inverse of x in S.

Answer 45

we should use \(\textbf{x}\) when we talk about arbitrary elements in \(S\) and \(x\) for arbitrary elements in \(\mathbb{R}\)

Answer 46

even that would get confusing... lol

Answer 47

too much info, trying to get in but i think I am getting it. and my confusion is increasing too. But dont worry abt exaplanation I am only getting more confused haha. Thanks a lot guys

Answer 48

Maybe the 1 is a multiplicative identity. In my course it was always seen as being the regular scalar 1 o.o Maybe we saw more restricted vector spaces?

Answer 49

correct @kirbykirby the 1 in the definition of vector space is referring to the multiplicative identity, which is often not the multiplicative identity in the vector space \(\mathbb{R}\)

Answer 50

wow I feel like I got cheated in my linear algebra course :( I don;t think they ever specified that, or just coincidentally all our vector space proofs involved the scalar 1 actually being the multiplicative inverse >_>

Answer 51

thanks for letting me know

Answer 52

identity*

Answer 53

You just mostly worked with normal R multiplication, this is more about the operations being different. Normally in linear algebra you use normal vector addition and scalar multiplication i.e. the normal real multiplication of the components of the vectors (which again are often real valued components, but need not be).