Kenneth Kuttler

402 APPENDIX B. CLASSIFICATION OF REAL NUMBERS

with αkvkxmk + · · ·+αkuk, it follows that for each large prime p,(αkv)p(mk+1) (αku)p =(α

mk+2k vmk+1

)p=U p. From now on, use the new Qk (x).

Defining fk (x) and Ik (s) as in Lemma B.3.4,

fk (x)≡v(m+1)pQp

k (x)xp−1

(p−1)!

and as before, let p be a very large prime number. It follows from Lemma B.3.4 that foreach k = 1, · · · ,n,

∑i=1

Ik (β (k)i) =

(Kk + ck

∑i=1

eβ (k)i

)deg( fk)

∑j=0

f ( j)k (0)

−

(Kk

deg( fk)

∑j=0

f ( j)k (0)+ ck

∑i=1

deg( fk)

∑j=0

f ( j)k (β (k)i)

)

This is exactly the same computation as in the beginning of that lemma except one addsand subtracts Kk ∑

deg( fk)j=0 f ( j)

k (0) rather than K ∑deg( fk)j=0 f ( j)

k (0) where the Kk are chosen suchthat their sum equals K and the term on the left converges to 0 as p→∞. By Lemma B.3.4,

∑i=1

Ik (β (k)i) =

(Kk + ck

∑i=1

eβ (k)i

)(U p +Nk p)

−Kk (U p +Nk p)− ckN′k p

(Kk + ck

∑i=1

eβ (k)i

)U p−KU p +Mk p

where Mk is some integer. Now add.

∑k=1

∑i=1

Ik (β (k)i) =U p

(K +

∑k=1

∑i=1

eβ (k)i

)−KmU p +Mp

If K +∑mk=1 ck ∑

mki=1 eβ (k)i = 0, then if p > max(K,m,U) you would have −KmU p +Mp

an integer so it cannot equal the left side which will be small if p is large. Therefore, ∗follows. ■

Next is an even more interesting Lemma which follows from the above corollary.

Lemma B.3.7 If b0,b1, · · · ,bn are non zero integers, and γ1, · · · ,γn are distinct alge-braic numbers, then

b0eγ0 +b1eγ1 + · · ·+bneγn ̸= 0

Proof: Assumeb0eγ0 +b1eγ1 + · · ·+bneγn = 0 (2.12)

Divide by eγ0 and letting K = b0,

K +b1eα(1)+ · · ·+bneα(n) = 0 (2.13)

402 APPENDIX B. CLASSIFICATION OF REAL NUMBERSwith ayvyx +-+-+ Q;u,z, it follows that for each large prime p, (arv)Pomer) (au)? =me+2 mg +1 Pka = U?. From now on, use the new Q, (x).Defining f; (x) and J; (s) as in Lemma B.3.4,yim)? (x) xP!(p—1)!and as before, let p be a very large prime number. It follows from Lemma B.3.4 that foreach k = 1,--- ,n,fx (x) =Lmk my deg (fx) (j)Ck Vk (B(k);) = [Kova ere) y fc!’ (0)i=l i=| jdeg (Sx) (j) my deg( fx) (j)_ [« y? fy (O)+ay y i! (B 0)j i=l j=0j=0This is exactly the same computation as in the beginning of that lemma except one addsand subtracts K; yee i) fe ) (0) rather than K yee ‘) fi ) (0) where the K; are chosen suchthat their sum equals K and the term on the left converges to 0 as p — ce. By Lemma B.3.4,Ck yh (B (k);) = [« + Ck y) (U? +Nxp)i=1 /i=1—K,(U? + Np) —cnNypmy= [cra he) U? —KU? +Mxpi=1i=where M,; is some integer. Now add.mYadk (B (k);) =U? [x Eafe) —KmU? +Mpil k=l i=lk=1If K+ ce Dey eP()i — 0, then if p > max(K,m,U) you would have —KmU? + Mpan integer so it cannot equal the left side which will be small if p is large. Therefore, «follows.Next is an even more interesting Lemma which follows from the above corollary.Lemma B.3.7 If bo,b1,--+ bn are non zero integers, and Y,,-++ ,Y, are distinct alge-braic numbers, thenboe” + bye" +---+bne™ £0Proof: Assumeboe” + bye! +---+b,e™ =0 (2.12)Divide by e% and letting K = bo,K+ bye) 4. +d, =0 (2.13)