Question

Proof of gauss lemma

Answer 1

@mukushla @Algebraic! @UnkleRhaukus @jim_thompson5910

any of you?

Answer 2

Well, hartnn I think you don't know exactly the meaning of "gauss lemma" ...should I tell you? that "may" help you in proving this

Answer 3

What level math is this?

Answer 4

i doubt that,but yes...

Answer 5

let there be a polynomial : \(P(x)\) be a monic polynomial of degree 'n' with coefficients over \(\mathbb{Z}\) (the set of integers). Suppose P(b) = 0 for some rational number 'b' then b 'must' be an integer

Answer 6

Ahhh, So much nerd talk!

Answer 7

@hartnn that is 'gauss lemma' can u now prove that?

Answer 8

i can try, but this would be my 1st time.

Answer 9

k make a try hartnn

Answer 10

any1?

Answer 11

I suck too much for this.

Answer 12

Can you just give me 5min to write the proof a bit formal.

Answer 13

Gauss Lema:
Let's assume b is rational, but not an integer => b=p/q where (p,q)=1 and q is not +/-1.
\[P(x) = x ^{n} + \sum_{0}^{n-1} x ^{i}a ^{i}\] so \[P(b)=P(\frac{ p }{ q }) = (\frac{ p }{ q }) ^{n} + \sum_{0}^{n-1} \frac{ p^{i}a^{i} }{ q^{i} }=0\] Multiply the whole thing by q^n we get: \[0 = p^{n} + \sum_{0}^{n-1} p^{i}q^{n-i}a^{i}\] Since i varries from 0 to n-1 it follows that n-i > 0 for each i so q is a divisor of the whole sum. Since q/0 follows that q/p^n but by definition (p,q) = 1 and q is not +/-1 => Contradiction.
=> Either b is an integer or b is not rational.

Answer 14

this is same but in other words
http://openstudy.com/study#/updates/5053d9e0e4b02986d3706eaf

Answer 15

can't it be done like this :

let \(P(x) = X^n + a_1 X^{n-1} + a_2 X^{n-2} + a_3 x^{n-3} +......+ a_n\) has a rational root *b* . Therefore substituting b in \(P(x) = 0\) , we get :

\(P(b) = b^n+a_1 b^{n-1}+a_2 b^{n-2} +...+a_n=0\)

\(b^n = -a_1 b^{n-1}-a_2b^{n-2} -....-a^n\)

since : \( b = p_1 ^{n_1} p_2 ^{n_2} ... p_k ^{n_k}\)

hence we can equate that as :

\((p_1 ^{n_1} p_2 ^{n_2} ... p_k ^{n_k})^n = -a_1 b^{n-1}-a_2b^{n-2} -....-a^n\)

Notice that the power of \(p_1\) on LHS is \(n*n_1\) . If \(n_1<0\), then essentially the power of \(p_1\) which can divide RHS? It can be almost \(n_1 *{n-1}\), because all the coefficients are integers and can only contribute positive powers of \(p_1\) , not negative powers. Hence \(n_1 \textbf{can not be negative}\) ...

Thus b is an integer ...

Answer 16

very nic

Answer 17

but, when u wrote b as a prime factorization u r already assuming that b is an integer. thus assumign what u r supposed to prove.

Answer 18

not yet until we prove that :

\(\large{ n_i > 0 }\)

Answer 19

does that make sense?

Answer 20

but you wrote:
"since: \(b=p_1^{n_1}\dotsb p_k^{n_k}\)"
this says that b is an integer.

Answer 21

Let \(p_1=2\)
and \(n_1=-l\)

Answer 22

perhaps if u add "where not all the \(n_i\)'s are necessarily nonnegative" and then prove that all these are actually nonnegative....

Answer 23

there is n_1 = 1

Answer 24

and hence UNTIL it is proved that n_i is positive we can't say that b>0

Answer 25

oops I meant b is an integer

Answer 26

@helder_edwin did that make sense?

Answer 27

yes that makes sense

Answer 28

u have this equation
\[ \large p_1^{n\times n_1}\dotsb p_k^{n\times n_k}=-a_1b^{n-1}-\dots-a_{n-1}b-a_n \]

Answer 29

yep

Answer 30

the LHS is the prime factorization (or looks like it) of the RHS

Answer 31

it looks a little bit tautological.

it makes no sense factoring p_i from the RHS because it seems that u already did it.

am i making any sense

Answer 32

wait for to minutes

Answer 33

brb

Answer 34

ok

Answer 35

I am back

Answer 36

hi. still here

Answer 37

i was reading your work. did u notice that u didn't use the fact that \(b\in\mathbb{Q}\)

Answer 38

I think I used: P(x) has a rational root b, P(b) =0

Answer 39

well talk in later

Answer 40

u r saying b is rational, but not using it

Answer 41

I will talk u later Please sorry holder

Answer 42

I meant helder

Answer 43

no problem. maybe tomorrow. here, it's quite late.

Answer 44

Please don't take negative, I am using pen tablet to write hence mistakes done

Answer 45

don't worry.