Question

Prove that any odd integer is of the form of the 4q+1, or 4q+3 where q is some integer

Answer 1

hope it's clear......

Answer 2

$\underline{\red{\bold{SOLUTION}}}$ ➡

$\bold{Let,} \\ \\ \bold{a \: \: be \: \: any \: \: positive \: \: number \: \: and \: \: b = 4} \\ \\ \bold{Then, \: \: by \: \: Euclid's \: \: Algorithim, \: \: We \: \: get,} \\ \\ \bold{a = 4q + r \:, \: \: for \: \: some \: \: integers \: q \geqslant 0} \\ \\ \bold{and \: \: r = 0 \: 1 \: 2 \: 3\: ,\: because \: \: 0 \leqslant r < 4 \: .} \\ \\ \bold{Therefore,} \\ \\ \bold{a = 4q \: \: or \: \: a = 4q + 1 \: \: or \: \: a = 4q + 2 \: \: or } \\ \bold{a = 4q + 3 \: .} \\ \\ \bold{Now,} \\ \\ \bold{a = 4q + 1} \\ \bold{ = > a = 2.(2q) + 1 } \\ \bold{ = > a= 2z + 1\: \: ;\: \: \: \: Where, \: \: z = 2q \: is \: \: an \: \: integer.} \\ \\ \bold{a = 4q + 3} \\ \bold{ = > a = 4q + 2 + 1} \\ \bold{ = > a = 2(2q + 1) + 1} \\ \bold{ = > a = 2z _ 1 + 1 \: \: ;\: \: \: \: Where ,\: \: z_1 = 2q + 1} \\ \\ \bold{Clearly, \: \: a = 4 q + 1 \: \: a = 4 q+ 3 \: \: are \: \: of \: \: the \: \:form } \\ \bold{of \: \: 2m + 1 ,\: \: where \: \: m \: \: is \: \: an \: \: integer.} \\ \\ \bold{Therefore, \: \: 4 q+ 1 \: \: and \: 4 q+ 3 \: \: are \: \: not \: \: divisible} \\ \bold{by \: \: 2 \: .} \\ \\ \bold{So ,\: \: these \: \: expressions \: \: of \: \: numbers \: \: are \: \: not } \\ \bold{even \: \: numbers;\: \: i.e. ,\: \: these \: \: are \: \: odd \: \: numbers.} \\ \\ \bold{Thus ,\: \: any \: \: odd \: \: integers \: \: can \: \: be \: \: expressed} \\ \bold{in \: \: the \: \: form \: \: of \: \: 4q + 1 \: \:4 q + 3 \: .}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold { \underline{Hence \: \: Proved.}}$

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✝ $\mathfrak{\red{THANKS...}}$ ✝