::: question
Question 1. Proof that
1ā+2ā+3āāÆ+nā for n>2 is
irrational.
:::
However, if we consider the linear combination of 1ā,
2ā, 3āā¦nā over Q, then it is
likely that we will get a similar result as we kick out some special
cases.
::: question
Question 2. Given n positive integers, a1ā<a2ā<āÆ<anā,
such that aiāajāāāā/Q for all i<j
and aiāāā/Q for all i. Also given
Ī»1ā,Ī»2āā¦Ī»nāāQ. Prove that
āi=1nāĪ»iāaiāāāQ if and only if
Ī»1ā=Ī»2ā=āÆ=Ī»nā=0.
:::
::: proof
Proof. Suppose this stand for n=k, then it is suffice to prove
n=k+1.
Suppose there exist
Ī»1ā,Ī»2āā¦Ī»k+1āāQ such that
āi=1k+1āĪ»iāaiāā=qāQ
If Ī»jā=0, then
āĪ»jāajāā+āi=1k+1āĪ»iāaiāā=qāQ,
and we are done according to the assumption.
If Ī»jāī =0 for all j, then we have
āi=1kāĪ»iāaiāā=qāĪ»k+1āak+1āā
Clearly, qāĪ»k+1āak+1āā is a root of a irreducible
quadratic polynomial gāQ[x]. And the polynomial must have
another root which is q+Ī»k+1āak+1ā.
Also, the polynomial
f=ā(xā(Ā±Ī»1āa1āā+Ā±Ī»2āa2āā+āÆ+Ā±Ī»nāanāā))
is also in Q[x] by induction.
The only thing left to proof is that ak+1āaiā,iā¤k satisfy the
assumption in the question, and then the induction assumption will
apply, which proves that Ī»iā=0,iā¤k.
As, we does not use any special property of Q other then the
fact that it is a field. So, the proof should be valid for any field
F that characteristic equals 0, if we rephrase the question
in term of F.
::: question
Question 3. Given n distinct element
a1ā,a2ā,ā¦,anāāF, let biā be a root of
x2āaiā for all i. And biābjā1āā/F for
all iī =j, and biāā/F for all i. Then given
Ī»1ā,Ī»2āā¦Ī»nāāF,
āi=0nāĪ»iābiāāF if and only if
Ī»iā=0 for all i.
:::
If we try to shift bkā to bkā+ckā such that
ckāāF, then the result clearly still stand. So, the
previous question can be generalised a bit.
::: question
Question 4. Given n distinct irreducible quadratic polynomial
f1ā,f2āā¦fnā in F[x].
And given any iī =j, fjā does not have root in
F[x]/fiā.
And biā be all roots of fiā for all i. Then given
Ī»1ā,Ī»2āā¦Ī»nāāF,
āi=0nāĪ»iābiāāF if and only if
Ī»iā=0 for all i.
:::
If we are not satisfied with the quadratic polynomials, we can try to
generalise the question to any polynomial. However, we may need to have
a more strict condition on the roots.
::: question
Question 5. Given n distinct irreducible polynomial
f1ā,f2āā¦fnā in F[x] with orders greater or
equal than 2.
Given any iī =j. fjā does not have root in
F[x]/fiā.
And biā be all roots of fiā for all i. Then given
Ī»1ā,Ī»2āā¦Ī»nāāF,
āi=0nāĪ»iābi,1āāF if and only if
Ī»iā=0 for all i.
:::
To prove the above question, we need another definition and some
relating lemma.
::: definition
Definition 1. Given a polynomial f(x)āF[x],
f(x)=xn+a1āxnā1+a2āxnā2+āÆ+anā1āx+anā
Define the derivative fā²(x) of f(x) as
fā²(x)=nxnā1+(nā1)a1āxnā2+(nā2)a2āxnā3+āÆ+2anā2āx+anā1ā
:::
Through some simple calculation, we can prove the following lemma.
::: lemma
Lemma 1. Given polynomials f(x),g(x)āF[x],
(f(x)g(x))ā²=fā²(x)g(x)+f(x)gā²(x)
:::
::: lemma
Lemma 2. Given a irreducible polynomial f(x)āF[x],
and a field K, such that FāK
and f(x) can be factorized into product of linear factors in
K.
Chose a root Ī± of f(x) in K. Then (xāĪ±)2
divides f(x) in K, if and only if fā²(x)=0.
:::
As fā²(x) is still a polynomial in F[x], and we can choose
arbitrary K. This lemma implies that irreducible polynomial
f(x) have distinct roots if and only if fā²(x)ī =0.
::: proof
Proof. If (xāĪ±)2 divides f(x) in K.
Let, f(x)=(xāĪ±)2g(x)
Then, fā²(x)=2(xāĪ±)g(x)+(xāĪ±)2gā²(x)
Then fā²(Ī±)=0. As f(x) is irreducible, and f(x) and fā²(x)
have common root. So, f(x) divides fā²(x).
As, the degree of fā²(x) is less than f(x), then fā²(x)=0.
Conversely, if fā²(x)=0.
Let, f(x)=(xāĪ±)g(x)
Then, fā²(x)=g(x)+(xāĪ±)gā²(x)
Then,
0ā=fā²(Ī±)=g(Ī±)+(Ī±āĪ±)gā²(Ī±)=g(Ī±)ā
So, Ī± is a root of g(x).
Thus, (xāĪ±)2 divides f(x) in K.Ā ā»
:::
::: lemma
Lemma 3. Given distinct irreducible polynomials
f(x),g(x)āF[x], where the characteristic of F
equals 0.
Chose any field K, such that
FāK, and f(x) and g(x) can be
factorized into product of linear factors in K.
Let a1ā,a2āā¦anā be all roots of f(x) in K,
and b1ā,b2āā¦bmā be all roots of g(x) in K.
Then, h(x)=āi=1nāāj=1mā(xā(aiā+bjā)) have
coefficients in F,
:::