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યુક્લિડની ભાગવિધિ અને વાસ્તવિક સંખ્યાઓ

સાબિત કરો કે, પ્રત્યેક પ્રાકૃતિક સંખ્યાને 5 k અથવા

bold 5 bold k bold space bold plus-or-minus bold space bold 1

અથવા

bold 5 bold k bold space bold plus-or-minus bold space bold 2 bold comma bold space bold space bold space bold k bold space bold element of bold n bold space bold union bold left curly bracket bold space bold 0 bold space bold right curly bracket

સ્વરૂપમાં દર્શાવી શકાય છે.

યુક્લિડના ભાગાકારના પુર્વપ્રમેય પરથી પ્રત્યેક પ્રાકૃતિક સંખ્યા a માટે b = 5  લેતાં અનન્ય અનૃણ પૂર્ણાંકો k અને r એવા મળે કે જેથી

$bold a bold space bold equals bold space bold 5 bold k bold space bold plus bold space bold r bold comma bold space bold 0 bold space bold less-than or slanted equal to bold space bold r bold space bold less than bold space bold 5$

પરંતુ $bold 0 bold space bold less-than or slanted equal to bold space bold r bold space bold less than bold space bold 5$ એટલે કે, r = 0, 1, 2, 3  અથવા 4

$bold therefore bold space bold a bold space bold equals bold space bold 5 bold k bold comma bold space bold a bold space bold equals bold space bold 5 bold k bold space bold plus bold space bold 1 bold comma bold space bold a bold space bold equals bold space bold 5 bold k bold space bold plus bold 2 bold comma bold space$

a = 5k + 3 અથવા a  સ્વરૂપની થાય.

જ્યાં, $bold K bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket$

હવે, a = 5k + 3

$bold equals bold space bold 5 bold k bold space bold plus bold space bold 5 bold space bold minus bold space bold space bold 2 bold equals bold space bold space bold 5 bold left parenthesis bold k bold space bold plus bold space bold 1 bold space bold right parenthesis bold minus bold 2 bold equals bold space bold 5 bold k bold apostrophe bold space bold minus bold space bold 2 bold comma bold space bold જ ્ ય ાં bold comma bold space bold k bold apostrophe bold space bold space bold equals bold space bold k bold space bold plus bold space bold 1 bold comma bold space bold k bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket bold equals bold space bold 5 bold k bold space bold minus bold space bold 2 bold comma bold space bold k bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket bold space bold space bold left parenthesis bold space bold because bold space bold ન ા bold space bold બદલ ે bold space bold k bold space bold લ ે ત ાં bold space bold right parenthesis bold space bold space$

$bold therefore bold space bold a bold space bold equals bold space bold 5 bold k bold space bold plus bold space bold 3 bold space$ ને બદલે a = 5k - 2 લખી શકાય.

તથા a = 5k + 4

$bold equals bold space bold 5 bold k bold space bold plus bold space bold 5 bold space bold minus bold space bold 1 bold equals bold space bold 5 bold left parenthesis bold k bold space bold plus bold space bold 1 bold right parenthesis bold space bold minus bold space bold 1 bold equals bold space bold 5 bold k bold apostrophe bold space bold minus bold space bold 1 bold comma bold space bold જ ્ ય ાં bold space bold k bold apostrophe bold space bold equals bold space bold k bold space bold plus bold space bold 1 bold comma bold space bold k bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket bold equals bold space bold 5 bold k bold space bold minus bold space bold 1 bold comma bold space bold k bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket bold space bold space bold space bold space bold space bold space bold space bold space bold space bold left parenthesis bold space bold therefore bold space bold k bold apostrophe bold space bold space bold ન ા bold space bold બદલ ે bold space bold k bold space bold લ ે ત ાં bold space bold right parenthesis bold space$

$bold therefore bold space bold a bold space bold equals bold space bold 5 bold k bold space bold plus bold space bold 4 bold space bold ન ે bold space bold space bold બદલ ે bold space bold a bold space bold equals bold space bold 5 bold k bold space bold minus bold space bold 1 bold space bold લખ ી bold space bold space bold શક ા ય bold. bold space$

$therefore$ પ્રત્યેક પ્રાકૃતિક સંખ્યાને $bold 5 bold k bold comma bold space bold 5 bold k bold space bold plus-or-minus bold space bold 1$ અથવા $bold 5 bold k bold space bold plus-or-minus bold space bold 2 bold comma bold space$ $bold k bold space bold element of bold space bold N bold space bold union bold space bold left curly bracket bold space bold 0 bold space bold right curly bracket bold space$ સ્વરૂપમં દર્શાવી શકાય છે.

સાબિત કરો કે, 2a + 3b અને 9a + 5b માથી કોઈ એક 17 વડે વિભાજ્ય હોય તો બીજી અભિવ્યક્તિ પણ 17 વડે વિભાજ્ય છે; a, b $bold element of$ N.
(સુચન 4(2a + 3b) + 9a + 5b = 17a + 17b)

સાબિત કરો કે, 6 અને પૂર્ણાંક n માં કોઈ સામાન્ય અવયવ ન હોય, તો n² - 1 એ 6 વડે વિભાજ્ય છે.

સાબિત કરો કે, n ધન યુગ્મ પૂર્ણાંક હોય, તો (n(n + 1) (n +2) એ 24 વડે વિભાજ્ય છે.

સાબિત કરો કે, એ n⁴ + 4n² +11 એ 16 વડે વિભાજ્ય છે, જ્યાં n એ અયુગ્મ પૂર્ણાંક છે.

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યુક્લિડની ભાગવિધિ અને વાસ્તવિક સંખ્યાઓ

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