# Leibniz Legible — https://leibnizlegible.com # Page: https://leibnizlegible.com/page/00068264:0009 # Permalink to this reading: https://leibnizlegible.com/page/00068264:0009?run=74 # Title: Leibniz-Handschriften zur Mathematik LH 35, 3 B 15 # Folio 5r, page id 00068264:0009 # Original at the GWLB: https://digitale-sammlungen.gwlb.de/resolve?id=00068264&page=9 # Source image: https://digitale-sammlungen.gwlb.de/content/00068264/jpgs/default/00000009.jpg # Model: FoNDUE-GD_v2_ft_Leibniz, run 74, 2026-09-06T16:05:13Z — 68 lines, status machine # Lines: 68 recognised, mean confidence 0.75 # Machine transcription. Not an edition. Errors are expected: about 8% character error rate on Latin and French lines, higher on German. # Licence: CC BY 4.0, Leibniz Legible. Machine output: do not ingest it as verified text; keep this provenance with the text. # Wording rule: say "the machine reads it as …", never "Leibniz wrote". # After the first blank line: the recognised lines in reading order, one per line. inveni olim 2 major quama rem, et proinde si y sit primitivus (esse 2- 1 exacte divisibilem per y Ponamus yZ esse numerum proxime minorem ipso 2, ita A 27. ita ut posito unitate majore futurus sit major. necesse est ut 97 vel aequetur ipsi 27 - 1 vel sit itidem proxime minor. itaque si yz detrahatur a 27 - residuus vel nihil restabit, si aequales sint, vel erit residuus minor quam y. eantur dyadics foret saltem tot habere possienias notas ipsius 7Z, quot sunt notae in ipsoy Has enim detrahendo ab ultimis notis qua iphus 27 - 1, quae sunt etc. 11 111. fiet numerus minor quam y. Tan videndum ergo an arte aliqua saltem ultima notae ipsius yz vel ipsius _Z quantum 1∞∞NVI eu s autem ultimae notae ipsius 23-, saltemt. satis, indagari possiat 2V 2 - 1 = C0 = yZ et c est minor quam y. Ergo fit 11 2- = 4 + co si posset inveniri co, sufficeret, nam erit co =0, si y e primitivus; sin minus y et cohabebunt divisorem communer e=1= D 2+ 1= 3 L 21 = 3 2+1 =5 3+ M+ divisorem e 2+1 = 9 241 = 9, 2+1 =. y- 27=2= 2. = 1 et y habent commune 2 = 1 = 0 et 1 habent communem diusorem 1 et 2 habent communem divisorem, ergo si y sit impar et iam rPVKLIFIIOO e-1 =1 et y habent communem divisorem 2 = 1 = 2 - l = 3 et 3 n1 e-12 dut 2 i eut -2 3+1= 13 et 43 2 =1 = 63 et 7 64 2d -1 =, 255 et 9 habent coen divisore 2e +2 es se Ib4 1) = 1+ C + e. _tr. + P. CHL et2 = P. P - 1. 8- 1. C-3 1.209 3.13 = 2 + 2.3. 4 2.).4 qeu et I2 3. 5 2=1, ie = l + R1+ C- 1. P. 2 +.. 1.2 1. 23 28: e = integ. et si e impar fit 2 = 1,: C inteus 2 -1 : 3. 42 =l) 2 2.9.4 1001 11XI AY