Hanyoyin Tabbatar da Lissafi
Hujja a fannin lissafi ita ce ginshiƙin wannan fanni. Hanyoyin tabbatarwa su ne ginshiƙin tabbatar da gaskiyar bayanin lissafi. Daga zato na asali zuwa ƙarshe, dole ne a tabbatar da cewa kowane mataki ya zama ingantacce. Fahimtar hanyoyin tabbatarwa daban-daban ba wai kawai yana ƙarfafa ƙwarewar nazari ba, har ma yana ƙara wa ƙwarewar koyo da amfani da lissafi a fannoni daban-daban.
Wannan labarin zai tattauna wasu daga cikin manyan hanyoyin tabbatarwa a fannin lissafi, gami da hujja kai tsaye, hujja a kaikaice (matsala da sabani), shigar da lissafi, da kuma shaida ta hanyar takamaiman misali. Kowace hanya tana da aikace-aikace daban-daban, ƙarfi, da rauni. Bari mu bincika su sosai.
1. Shaida Kai Tsaye
Ma'ana da Misalai
Hujja kai tsaye hanya ce da muke tabbatar da magana ta hanyar nuna cewa idan hujjoji (zato) gaskiya ne, to ƙarshen ma gaskiya ne. A cikin hujja kai tsaye, yawanci muna farawa da abin da aka sani kuma muna amfani da matakai masu ma'ana don cimma ƙarshe.
Misali:
Ka tabbatar da cewa idan \(n\) lamba ce mai daidaito, to \(n^2\) ma daidai take.
Shaida:
A ce \(n\) lamba ce mai daidaito. To, bisa ga ma'anar lamba mai daidaito, ana iya rubuta cewa \(n = 2k\) ga wani lamba \(k\). Don haka,
\[n^2 = (2k)^2 = 4k^2 = 2 (2k^2) \]
A bayyane yake cewa \(n^2\) za a iya bayyana shi sau 2 a lamba ɗaya (watau \(2k^2\)). Tunda babban abin da ake buƙata don lamba mai daidaito shine cewa ana iya bayyana shi sau 2 a lamba ɗaya, to \(n^2\) shi ma lamba ce mai daidaito.
2. Shaida ta Kai Tsaye
Hujja ta kai tsaye ta ƙunshi manyan hanyoyi guda biyu: hujja ta hanyar sabani da kuma hujja ta hanyar sabani.
a. Shaidar Sabani
Ma'ana da Misalai
Wannan hanyar ta ƙunshi tabbatar da bayanin da ke da alaƙa da "idan \(P\), to \(Q\)" ta hanyar tabbatar da sabanin bayanin: "idan ba \(Q\), to ba \(P\)" ba.
Misali:
Ka tabbatar da cewa idan \(n^2\) baƙo ne, to \(n\) shi ma baƙo ne.
Shaida:
Sabanin wannan magana shine: Idan \(n\) ba mai ban mamaki ba ne (ko ma mai ban mamaki), to \(n^2\) ba mai ban mamaki ba ne (ko ma mai ban mamaki).
A ce \(n\) daidai ne, to \(n = 2k\) ga lamba \(k\). Don haka,
\[n^2 = (2k)^2 = 4k^2 = 2 (2k^2) \]
Wannan yana nufin cewa \(n^2\) lamba ce mai daidaito. Don haka, an tabbatar da abin da ya saba wa juna, kuma an tabbatar da cewa asalin maganar gaskiya ce.
b. Shaida ta Sabani
Ma'ana da Misalai
Shaida ta hanyar sabani ta ƙunshi ɗauka cewa maganar da za a tabbatar ƙarya ce da kuma nuna cewa wannan zato yana haifar da sabani mai ma'ana.
Misali:
Ka tabbatar da cewa \(\sqrt{2}\) lamba ce mara ma'ana.
Shaida:
A ce, maimakon haka, cewa \(\sqrt{2}\) lamba ce mai ma'ana. Sannan, \(\sqrt{2} = \frac{a}{b}\), inda \(a\) da \(b\) su ne manyan lambobi (ragewa shine 1), da kuma \(b \ne 0\). Don haka, za mu iya rubutawa:
\[ \sqrt{2} = \frac{a}{b} \]
\[ 2 = \frac{a^2}{b^2} \]
\[ 2b^2 = a^2 \]
Daga wannan lissafi, mun ga cewa \(a^2\) lamba ce mai daidaito, wanda ke nufin \(a\) dole ne ta kasance daidai. A ce \(a = 2k\), muna da:
\[ 2b^2 = (2k)^2 \]
\[ 2b^2 = 4k^2 \]
\[ b^2 = 2k^2 \]
Tunda \(b^2\) lamba ce mai daidaito, to \(b\) dole ne ta kasance lamba mai daidaito. Wannan yana nufin cewa \(a\) da \(b\) duka lambobi ne masu daidaito, wanda ya saɓa wa zato na asali cewa \(\frac{a}{b}\) yana cikin mafi sauƙin siffa. Saboda haka, \(\sqrt{2}\) ba zai iya zama lamba mai hankali ba, saboda haka ba shi da hankali.
3. Shigar da Lissafi
Ma'ana da Misalai
Injin lissafi hanya ce ta tabbatar da bayanai da suka shafi lambobi. Tsarin ya ƙunshi matakai biyu: tushen shigarwa da kuma matakin shigarwa.
Misali:
Tabbatar da cewa jimlar jerin lambobi na farko \(1 + 2 + 3 + … + n = \frac{n(n+1)}{2}\).
Shaida:
- Tushen Shigarwa:
Domin \(n = 1\),
\[ 1 = \frac{1(1+1)}{2} \]
daidai.
- Matakan Shigarwa:
A ɗauka cewa wannan magana gaskiya ce ga lamba \(k\). Wato,
\[ 1 + 2 + 3 + … + k = \frac{k(k+1)}{2} \]
Muna buƙatar tabbatar da cewa hakan ma gaskiya ne ga \(k + 1\). Mun ƙara \((k + 1)\) zuwa ɓangarorin biyu na lissafin:
\[ 1 + 2 + 3 + … + k + (k + 1) = \frac{k(k+1)}{2} + (k + 1) \]
\[ = \frac{k(k+1) + 2(k+1)}{2} \]
\[ = \frac{(k + 1)(k + 2)}{2} \]
Don haka, maganar gaskiya ce ga \(k + 1\). Saboda haka, bisa ga ka'idar induction na lissafi, maganar gaskiya ce ga dukkan lambobi masu kyau \(n\).
4. Shaida tare da Takamaiman Misalai
Ma'ana da Misalai
Wannan hanyar ta ƙunshi tabbatarwa ta hanyar zaɓar takamaiman misalai waɗanda suka cika dukkan sharuɗɗan da aka bayar a cikin sanarwar kuma suna nuna cewa maganar gaskiya ce. Duk da haka, galibi ana amfani da wannan hanyar don tabbatar da ƙaryar magana.
Misali:
Ka tabbatar da cewa akwai lambobi da ba za a iya bayyana su a matsayin jimlar murabba'i biyu masu kyau ba.
Shaida:
Gwada amfani da misali \(3\):
A ce \(3\) za a iya bayyana shi a matsayin jimlar murabba'ai biyu masu kyau, wato \(a^2 + b^2 = 3\). Bayan gwada dukkan haɗuwar lambobi \(a\) da \(b\),
1. \(a = 0\), \(b^2 = 3\) (ba zai yiwu ba).
2. \(a = 1\), \(b^2 = 2\) (ba zai yiwu ba).
3. \(a = 2\), \(b^2 = -1\) (ba zai yiwu ba).
4. Lambobi ko lambobi masu kama da na korau ko kuma lambobi sama da 2 ba za su yiwu ba.
Wannan yana nuna cewa ba za a iya bayyana \(3\) a matsayin jimlar lambobi murabba'i biyu ba. Don haka, akwai lambobi waɗanda ba za a iya bayyana su a matsayin jimlar lambobi murabba'i biyu masu cikakke ba.
Kammalawa
Shaida a cikin lissafi tana buƙatar hanyoyi daban-daban da matakai na tsari dangane da nau'in bayanin da aka tabbatar. Shaida kai tsaye, shaida kai tsaye (masu karo da sabani), shigar da lissafi, da misalai na musamman wasu daga cikin manyan hanyoyin shaida da ake amfani da su a yanayi daban-daban. Fahimtar waɗannan hanyoyin zai ƙarfafa tushen lissafi kuma ya taimaka muku zurfafa bincike kan sassa daban-daban na lissafi.
Tare da aiki da fahimta mai zurfi, hanyoyin tabbatar da lissafi za su zama kayan aiki wanda koyaushe zai kasance a shirye don amfani da shi wajen magance matsalolin lissafi masu rikitarwa.
