proof of gauss lemna

Question

anonymous · Answer

anyone ?

anonymous · Answer

@sauravshakya @hartnn @zarkon

hartnn · Answer

did not learn number theory in that details......
@mukushla

anonymous · Answer

$$P(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0 \ \ \ \ \ \ a_i \in \mathbb{Z}$$suppose  $P(b)=0$ and  $b=\frac{p}{q}$ such that  $\gcd(p,q)=1$  and p,q are some integers$$b^n+a_{n-1}b^{n-1}+a_{n-2}b^{n-2}+...+a_1b+a_0=0$$$$(\frac{p}{q})^n+a_{n-1}(\frac{p}{q})^{n-1}+a_{n-2}(\frac{p}{q})^{n-2}+...+a_1(\frac{p}{q})+a_0=0$$multiply by  $q^n$ $$p^n+a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^2+...+a_1pq^{n-1}+a_0q^n=0$$$$a_{n-1}p^{n-1}q+a_{n-2}p^{n-2}q^2+...+a_1pq^{n-1}+a_0q^n=-p^n$$LHS is divisible by q so RHS must be divisible by q so$$q|-p^n$$but p and q are coprime so  $q=\pm1$  and  b is an integer

mathslover · Answer

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 ...

anonymous · Answer

emm..

b is rational so we cant write it in the form$$b = p_1 ^{n_1} p_2 ^{n_2} ... p_k ^{n_k}$$??

mathslover · Answer

We can write b as a product of prime, where the powers can be positive or negative for a rational number. ...

anonymous · Answer

oh so $n_i$  can be negative ok

mathslover · Answer

n_i >0

mathslover · Answer

It can't be negative ... n_ i >0 can be proved

anonymous · Answer

i mean firstly we can suppose that it can be negative then prove that it must be positive

mathslover · Answer

p_1 has positive power .... that is n_1 > 0 

similarly ... n_i>0

mathslover · Answer

oh k

mathslover · Answer

"contradiction" ..  ? right.. ?

anonymous · Answer

beautiful

mathslover · Answer

thanks... :)

mathslover · Answer

I loved this proof .... thanks a lot mukushla

anonymous · Answer

np :)

mayankdevnani · Answer

attachment

mayankdevnani · Answer

@Kushashwa_ravi