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# Title: Leibniz-Handschriften zur Mathematik LH 35, 15, 5
# Folio 78v, page id 00068207:0156
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(5) = 4
p0 +
b Lnab 23 abc 424 abcd 425 abede, )-
2 22 pa + 23p ab 1 24 abc + 25 pabid + 26 abede )
= 22 ab - 23 abc - 24 ab cd - 25 abcoe
D) = p0 - D 5+ (D) = pC
 ita accipi debet, ut plus tribuatur majori
ex differentibus pC et D. vel certe cuilubet
Cum sunt aequalia
Erx A. Necesse est D + (D) esse afformativam
si d est 1, non potest +() esse +) sed vel )
D est 0, vel + est -
+ 22. Cab) + 23 ( abc) + R + sab c0) f. 23 sab cdC)
aequ.
pa - 2pab + e p abe
eituleteam
inveni etiam sequentia thecremata
D = 101 a. F 102 ab + 103 abc +104 abcd
(a) = a = p (ab) - ab = pa (abc)- abc = p ab et ita P
(8) = 101 (a) + 102 (ab) + 103 sabC)
Ergo ((B) - B,: p = 101. 1 + 102 a + 103 ab + 104 abc t etc.
O_t
(0) =  X  (D) = 9+ O p 20dp
O2H
 diff. C et p ducta in, D, 1- 2D
P -C in pO =0. Ergo O+ p - 2) p in pO =0
C - p, in p0 = 0 Ergo
Ergo O+p -20p, in D + Op (1- 2D) = Orp . 26p in D
Erp (C)D) = (0) D
seo non ideo
3) =
Op 2p in (D)
Ermo (0D) - CD, : p = Bs () D - D
Eeb +a +b in ab ii aa = a et ab bb - b
uut qab. L ab itaque fa + b, in ab =0, licet
enim nunquam ab s non vil ab, non ideo sit 0,
nec ab, ideo sit, 0, necessario tamen vel ab est 0,
vel + a + b est 0. unde productum fit 0
Htinc si (1() = (C) D non tequitur esse 5) esse (5)
sed hoc sequitur, vet C esse 0, vet D esse (D)
(D) est differentia inter D et pC
(D) = +5 Y D+ p0
(() est differentia inter C et p
(O = 1 a +in + 13 abc + 1+ abcd
(6) = 11 (a) + 12 (al) + 13(abc) + 14 (abc d) = + n a + 12 ab + B abe tera
a) = a + P (ab) = ab +pa (abc) = ab e+t pat (abcd = abed + pabeterc
aD
ti + na +12ab + 13 abc + 14 abcd + CtC
1p +12pa + p ab + 14p abC 15 pabed
et fiet 11.
20
+p Qua + nab Hi3abe f 14 abid
quae aequatio ientica, ergo 1141 =0 seu 1 - + 1 Ergo squadrando
fit  B= sed alliunde scimus esse H 1 ergo fiet
+C  1 Et + si p = 1 fit 11 = + 1 debent
autem veri numeri provenire, etiam cum p=1, quia sunt semper
sidem jum scimus et si p = 0 nihil ex calculo dum
potest. Porro scimus aliunde esse  11 = + 1 ergo
1 et aequatio erit
tia + 12ab + 13abc + etC)
fit 4 =
HIy
+H +1a + 1ab + igabe
-1 + Ha 4 12 ab + 13 abc
sea +n + 1a + 13 ab + 14 abc ) = C
1 - 2. Ha = 2 12ab = 2. (3abc 5
Undetandem 11 = 1 1 = - 2 13- + 4, 1 - - 8 et
Mlurcto
vel pesto.
vel a esto
sed 22 non est 0
Ergo vel p vel a
esto