Edukiak
Biderketako termino berdinen batura gisa irudika daitekeen eragiketa matematiko bat da.
Biderketaren printzipio orokorra
Adibidez, a ⋅ b («a aldiz b» bezala irakurrita) terminoak batutzen ditugula esan nahi du a, zeinaren kopurua berdina den b. Biderketa baten emaitzari produktu deritzo.
adibideak:
- 2 ⋅ 6 = 2 + 2 + 2 + 2 + 2 + 2 = 12
(sei aldiz bi)
- 5 ⋅ 4 = 5 + 5 + 5 + 5 = 20
(lau aldiz bost)
- 3 ⋅ 8 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 24
(zortzi aldiz hiru)
Dakigunez, faktoreen lekuen permutaziotik, produktua ez da aldatzen. Goiko adibideetarako, hauxe da:
- 6 ⋅ 2 = 6 + 6 = 12
(bi aldiz sei)
- 4 ⋅ 5 = 4 + 4 + 4 + 4 + 4 = 20
(bost aldiz lau)
- 8 ⋅ 3 = 8 + 8 + 8 = 24
(hiru aldiz zortzi)
Onura praktikoak
Biderketari esker, mota bereko elementu guztien kopurua nabarmen murriztu dezakezu, etab. Esate baterako, 7 pakete baditugu, bakoitzak 5 boligrafo dituena, boligrafoen kopuru osoa hauek biderkatuz aurkitzen da. bi zenbaki:
5 ⋅ 7 = 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35
(bost boligrafo zazpi aldiz)
Biderkatu 0
Emaitza beti zero da.
- 0 ⋅ 0 = 0
- 1 ⋅ 0 = 0 ⋅ 1 = 0
- 2 ⋅ 0 = 0 ⋅ 2 = 0 + 0 = 0
- 3 ⋅ 0 = 0 ⋅ 3 = 0 + 0 + 0 = 0
- 4 ⋅ 0 = 0 ⋅ 4 = 0 + 0 + 0 + 0 = 0
- 5 ⋅ 0 = 0 ⋅ 5 = 0 + 0 + 0 + 0 + 0 = 0
- 6 ⋅ 0 = 0 ⋅ 6 = 0 + 0 + 0 + 0 + 0 + 0 = 0
- 7 ⋅ 0 = 0 ⋅ 7 = 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0
- 8 ⋅ 0 = 0 ⋅ 8 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0
- 9 ⋅ 0 = 0 ⋅ 9 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0
- 10 ⋅ 0 = 0 ⋅ 10 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0
Biderkatu 1
Produktua bat ez den beste biderkatzaile baten berdina da.
- 1 ⋅ 1 = 1
- 2 ⋅ 1 = 2 ⋅ 1 = 2
- 3 ⋅ 1 = 3 ⋅ 1 = 3
- 4 ⋅ 1 = 4 ⋅ 1 = 4
- 5 ⋅ 1 = 5 ⋅ 1 = 5
- 6 ⋅ 1 = 6 ⋅ 1 = 6
- 7 ⋅ 1 = 7 ⋅ 1 = 7
- 8 ⋅ 1 = 8 ⋅ 1 = 8
- 9 ⋅ 1 = 9 ⋅ 1 = 9
- 10 ⋅ 1 = 10 ⋅ 1 = 10
Biderkatu 2
Gehitu lehen faktorea bere buruari.
- 1 ⋅ 2 = 1 + 1 = 2
- 2 ⋅ 2 = 2 + 2 = 4
- 3 ⋅ 2 = 3 + 3 = 6
- 4 ⋅ 2 = 4 + 4 = 8
- 5 ⋅ 2 = 5 + 5 = 10
- 6 ⋅ 2 = 6 + 6 = 12
- 7 ⋅ 2 = 7 + 7 = 14
- 8 ⋅ 2 = 8 + 8 = 16
- 9 ⋅ 2 = 9 + 9 = 18
- 10 ⋅ 2 = 10 + 10 = 20
Biderkatu 3
Lehenengo faktorea 2z biderkatzen dugu, ondoren emaitzari gehituko diogu.
- 1 ⋅ 3 = (1 ⋅ 2) + 1 = 2 + 1 = 3
- 2 ⋅ 3 = (2 ⋅ 2) + 2 = 4 + 2 = 6
- 3 ⋅ 3 = (3 ⋅ 2) + 3 = 6 + 3 = 9
- 4 ⋅ 3 = (4 ⋅ 2) + 4 = 8 + 4 = 12
- 5 ⋅ 3 = (5 ⋅ 2) + 5 = 10 + 5 = 15
- 6 ⋅ 3 = (6 ⋅ 2) + 6 = 12 + 6 = 18
- 7 ⋅ 3 = (7 ⋅ 2) + 7 = 14 + 7 = 21
- 8 ⋅ 3 = (8 ⋅ 2) + 8 = 16 + 8 = 24
- 9 ⋅ 3 = (9 ⋅ 2) + 9 = 18 + 9 = 27
- 10 ⋅ 3 = (10 ⋅ 2) + 10 = 20 + 10 = 30
Biderkatu 4
Bikoiztutako lehenengo faktoreari kopuru bera gehitzen diogu.
- 1 ⋅ 4 = (1 ⋅ 2) + (1 ⋅ 2) = 2 + 2 = 4
- 2 ⋅ 4 = (2 ⋅ 2) + (2 ⋅ 2) = 4 + 4 = 8
- 3 ⋅ 4 = (3 ⋅ 2) + (3 ⋅ 2) = 6 + 6 = 12
- 4 ⋅ 4 = (4 ⋅ 2) + (4 ⋅ 2) = 8 + 8 = 16
- 5 ⋅ 4 = (5 ⋅ 2) + (5 ⋅ 2) = 10 + 10 = 20
- 6 ⋅ 4 = (6 ⋅ 2) + (6 ⋅ 2) = 12 + 12 = 24
- 7 ⋅ 4 = (7 ⋅ 2) + (7 ⋅ 2) = 14 + 14 = 28
- 8 ⋅ 4 = (8 ⋅ 2) + (8 ⋅ 2) = 16 + 16 = 32
- 9 ⋅ 4 = (9 ⋅ 2) + (9 ⋅ 2) = 18 + 18 = 36
- 10 ⋅ 4 = (10 ⋅ 2) + (10 ⋅ 2) = 20 + 20 = 40
Biderkatu 5
Beste biderkatzailea zenbaki bikoitia bada, emaitza zeroan amaituko da, bakoitia bada, 5 zenbakian.
- 1 ⋅ 5 = 5 ⋅ 1 = 5
- 2 ⋅ 5 = 5 ⋅ 2 = 5 + 5 = 10
- 3 ⋅ 5 = 5 ⋅ 3 = (5 ⋅ 2) + 5 = 15
- 4 ⋅ 5 = 5 ⋅ 4 = (5 ⋅ 2) + (5 ⋅ 2) = 20
- 5 ⋅ 5 = 5 + 5 + 5 + 5 + 5 = 25
- 6 ⋅ 5 = 5 ⋅ 6 = (5 ⋅ 5) + 5 = 30
- 7 ⋅ 5 = 5 ⋅ 7 = 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35
- 8 ⋅ 5 = 5 ⋅ 8 = (5 ⋅ 4) + (5 ⋅ 4) = 40
- 9 ⋅ 5 = 5 ⋅ 9 = (5 ⋅ 10) – 5 = 45
- 10 ⋅ 5 = 5 ⋅ 10 = 50
Biderkatu 6
Lehenengo faktorea 5ez biderkatuko dugu, gero emaitza gehituz.
- 1 ⋅ 6 = (1 ⋅ 5) + 1 = 5 + 1 = 6
- 2 ⋅ 6 = (2 ⋅ 5) + 2 = 10 + 2 = 12
- 3 ⋅ 6 = (3 ⋅ 5) + 3 = 15 + 3 = 18
- 4 ⋅ 6 = (4 ⋅ 5) + 4 = 20 + 4 = 24
- 5 ⋅ 6 = (5 ⋅ 5) + 5 = 25 + 5 = 30
- 6 ⋅ 6 = (6 ⋅ 5) + 6 = 30 + 6 = 36
- 7 ⋅ 6 = (7 ⋅ 5) + 7 = 35 + 7 = 42
- 8 ⋅ 6 = (8 ⋅ 5) + 8 = 40 + 8 = 48
- 9 ⋅ 6 = (9 ⋅ 5) + 9 = 45 + 9 = 54
- 10 ⋅ 6 = (10 ⋅ 5) + 10 = 50 + 10 = 60
Biderkatu 7
Ez dago 7rekin biderkatzeko algoritmo sinplifikaturik, beraz, beste faktore batzuei aplikagarri zaizkien metodoak erabiltzen ditugu.
- 1 ⋅ 7 = 7 ⋅ 1 = 7
- 2 ⋅ 7 = 7 ⋅ 2 = 7 + 7 = 14
- 3 ⋅ 7 = 7 ⋅ 3 = (7 ⋅ 2) + 7 = 21
- 4 ⋅ 7 = 7 ⋅ 4 = (7 ⋅ 2) + (7 ⋅ 2) = 28
- 5 ⋅ 7 = 7 ⋅ 5 = 7 + 7 + 7 + 7 + 7 = 35
- 6 ⋅ 7 = 7 ⋅ 6 = (7 ⋅ 5) + 7 = 42
- 7 ⋅ 7 = 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49
- 8 ⋅ 7 = 7 ⋅ 8 = (7 ⋅ 4) + (7 ⋅ 4) = 56
- 9 ⋅ 7 = 7 ⋅ 9 = (7 ⋅ 10) – 7 = 63
- 10 ⋅ 7 = 70
Biderkatu 8
Lehenengo faktorea 4z biderkatuko dugu, eta emaitzari kopuru bera gehituko diogu.
- 1 ⋅ 8 = (1 ⋅ 4) + (1 ⋅ 4) = 8
- 2 ⋅ 8 = (2 ⋅ 4) + (2 ⋅ 4) = 16
- 3 ⋅ 8 = (3 ⋅ 4) + (3 ⋅ 4) = 24
- 4 ⋅ 8 = (4 ⋅ 4) + (4 ⋅ 4) = 32
- 5 ⋅ 8 = (5 ⋅ 4) + (5 ⋅ 4) = 40
- 6 ⋅ 8 = (6 ⋅ 4) + (6 ⋅ 4) = 48
- 7 ⋅ 8 = (7 ⋅ 4) + (7 ⋅ 4) = 56
- 8 ⋅ 8 = (8 ⋅ 4) + (8 ⋅ 4) = 64
- 9 ⋅ 8 = (9 ⋅ 4) + (9 ⋅ 4) = 72
- 10 ⋅ 8 = (10 ⋅ 4) + (10 ⋅ 4) = 80
Biderkatu 9
Lehenengo faktorea 10ez biderkatzen dugu, eta ondoren lortutako emaitzari kenduko diogu.
- 1 ⋅ 9 = (1 ⋅ 10) – 1 = 10 – 1 = 9
- 2 ⋅ 9 = (2 ⋅ 10) – 2 = 20 – 2 = 18
- 3 ⋅ 9 = (3 ⋅ 10) – 3 = 30 – 3 = 27
- 4 ⋅ 9 = (4 ⋅ 10) – 4 = 40 – 4 = 36
- 5 ⋅ 9 = (5 ⋅ 10) – 5 = 50 – 5 = 45
- 6 ⋅ 9 = (6 ⋅ 10) – 6 = 60 – 6 = 54
- 7 ⋅ 9 = (7 ⋅ 10) – 7 = 70 – 7 = 63
- 8 ⋅ 9 = (8 ⋅ 10) – 8 = 80 – 8 = 72
- 9 ⋅ 9 = (9 ⋅ 10) – 9 = 90 – 9 = 81
- 10 ⋅ 9 = (10 ⋅ 10) – 10 = 100 – 10 = 90
Biderkatu 10
Gehitu zero beste biderkatzailearen amaieran.
- 1 ⋅ 10 = 10 ⋅ 1 = 10
- 2 ⋅ 10 = 10 ⋅ 2 = 20
- 3 ⋅ 10 = 10 ⋅ 3 = 30
- 4 ⋅ 10 = 10 ⋅ 4 = 40
- 5 ⋅ 10 = 10 ⋅ 5 = 50
- 6 ⋅ 10 = 10 ⋅ 6 = 60
- 7 ⋅ 10 = 10 ⋅ 7 = 70
- 8 ⋅ 10 = 10 ⋅ 8 = 80
- 9 ⋅ 10 = 10 ⋅ 9 = 90
- 10 ⋅ 10 = 10 ⋅ 10 = 100