Question

proof that any arbitrary of "C" shouldbe included all of Rational numbers !
(linear Algebra)

Answer 1

Prove what? :/

Answer 2

N<Z<Q<R<C

Answer 3

I don't quite get it... any arbitrary <what> of "C" ?

Answer 4

sorry my english is not very good so i mean prove that any arbitrary subfield of "C" shouldbe included all of Rational numbers !

Answer 5

subfield :)

Answer 6

Ohh... subfields... okay
Isn't this more of an Abstract Algebra problem and not a Linear Algebra one? :D
Nonetheless... let's proceed...

Answer 7

@Diogo You wanna do this? :D

Answer 8

Oh...kay... @horotat
You ready to proceed?

Answer 9

Well.. All imaginary numbers come from a negative square root which is an irrational number.
Let's assume that \[\sqrt{2}\] is a rational number.

Now lets say that there are 2 rational numbers that when divided they give you sqrt(2)

\[\frac{ p }{ q } = \sqrt{2}\]
\[(\frac{ p }{ q })^{2} = 2\]
p^2= 2q^2

p is even so p^2 is even as well because an even number when multiplied by itself is even too.

so you can say that p=2k

(2k)^2 = 2q^2

4k^2 = 2q^2

2k^2 = q^2

And now you proved that q is also an even number...

But that doesnt even make sense because when you devide p/q you get sqrt(2)

which means sqrt(2) = irrational number.

So if an imaginary number come from an irrational number and an irrational number come from rational numbers then all rational numbers are a subfield of any arbitrary C

Answer 10

wow, thank u all .

Answer 11

Honestly @Diogo that made perfect sense, but it doesn't prove that any subfield of C must contain the set of Rationals...
My turn :)

Let F be a subfield of C.
We have to show that Q is contained in F.
In other words, all rational numbers would be in F.

Since F is a subfield, it must contain unity (1)
Since F is a subfield, it is also a group, hence the additive inverse of 1, which is -1, is also in F.

Since F is a field, then any finite sum of 1's (1+1+1...+1) would also be in F.
Similarly, any finite sum of -1's (-1-1-1...-1) would also be in F.

This shows that all integers are in F.

Since F is a field, then every nonzero element would have a multiplicative inverse, or something that you can multiply to it so that the product is unity (1).

So, we have shown that all integers are in F. Then let q be a rational number.
Then q = a/b, where a and b are integers.

a/b = ab^-1
And since a and b^-1 are in F, so is ab^-1 which is equal to a/b which is equal to q.

Thus, for any rational number q, q would be in F.
Thus, Q (the set of Rational numbers) would be contained in F.

Q(uite)
E(asily)
D(one)

LOL
I hope you guys enjoyed that :)

Answer 12

@Diogo thanks alot :)
& @terenzreignz THANK you more alot :D
openstudy is really relly great for me ...

Answer 13

Stick around, @horotat and see if you can be of help to others yourself :)
Welcome to OpenStudy, and enjoy :D

Answer 14

:) thanks again, i hope that i'll be helpful