Question

http://math.stackexchange.com/questions/766935/prove-that-if-y-z-in-q-then-yz-in-a

Answer 1

what is with that \[y^z \in A\] it's throwing me off.

Answer 2

hmm, is this for algebra class?

Answer 3

lol no. proof writing ^^

Answer 4

hmm, would be nicer with algebra theorems:) I got to go watch mad men with the wife but ill come back after its over, this seems fun....

Answer 5

if I have an irrational number, then I can't prove this because it will break a proposition. So since x and y belong to Q which is the set of rational numbers, then by the prop it must be algebraic. but w t fuu fuu is that y^z?!

Answer 6

I'm talking abstract algebra where its very common to study algebraic numbers with field extensions

Answer 7

D: I'm not even in that class...so idk what's going on.

Answer 8

you just need to find a polynomial that has a root of \((\frac{a}{b})^{\frac{c}{d}}\)

Answer 9

in \(\mathbb{Z[x]}\)

Answer 10

O_O that's gonna be a very long polynomial string isn't it?

Answer 11

oh lord it's going to be like that example... the one with If a in A and q in Q then qa in A

Answer 12

example proof:

let \[a \in A\] and \[q\in Q\]. By the definitions, there exists a polynomial \[p \in Z[x]\] of the form \[p(x) = c_nx^n+c_{n-1}x^{n-1}+...+c_1x+c_0 \]

Answer 13

such that p(a) = - and there exists integers k and l such that \[l \neq 0 \] and \[q = \frac{k}{l} \] Thus
\[c_na^n+c_{n-1}a^{n-1}+...+c_aa+c_o=0\]

Answer 14

which we multiply by \[q^n \] to obtain \[c_n(\frac{ak}{l})^n+c_{n_1}\frac{k}{l}(\frac{ak}{l})^{n-1} +...+c_1\frac{k^{n-1}}{l^{n-1}}(\frac{ak}{l})+\frac{k^n}{l^n}c_0 = 0\]
multiply the last equation by \[$l^n$\] to obtain
\[l^nc_n(aq)^n+l^{n-1}c_{n-1}k(aq)^{n-1}+...+lc_ak^{n-1}aq+k^nc_0=0\]

Answer 15

\[p_1(x) = l^nc_nx^n+l^{n-a}c_{n-1}kx^{n-1}=...=lc_1k^{n-1}x+k^nc_0\] also belongs to Z[x] . By equation 2.7.1 p_1(aq) = 0. Thus qa in A

Answer 16

@ganeshie8

Answer 17

@Kainui

Answer 18

since \(y, z\in \mathbb Q \),

say \(y = \dfrac{a}{b} , z = \dfrac{c}{d}\) such that \(a, b, c, d \in \mathbb Z\)

Answer 19

u are trying to prove \(y^z\) is a root to some polynomial, right ?

Answer 20

yeah.

Answer 21

I gotta find some polynomial with this root \[(\frac{a}{b})^{\frac{c}{d}} \]

Answer 22

in Z[x]

Answer 23

set it equal to \(x \)

Answer 24

\(y^p = x\)


\(\left(\dfrac{a}{b} \right)^{\dfrac{c}{d}}= x\)

Answer 25

take power "d" both sides :

\(\left(\dfrac{a}{b} \right)^{c}= x^d\)

Answer 26

multiply both sides by \(b^c\)

Answer 27

\(a^c= b^cx^d\)

Answer 28

\( b^cx^d - a^c = 0\) is the polynomial which has \(y^p\) as root.
QED right ?

Answer 29

yeah

Answer 30

someone on stackexchange arrived at that answer and I was trying to figure out how he got there.

Answer 31

so I thought about subsituting at first.

Answer 32

can you look at another question ^^?

Answer 33

http://math.stackexchange.com/questions/767117/let-a-b-in-r-where-a-b-prove-that-there-exist-a-rational-number-c-and/767138?noredirect=1#767138

Answer 34

I sort of got the idea but I'm not sure if I'm on the right track because I've read about rational/irrational numbers with decimal expansions