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Show that any positive odd integer is of the form, 6q+1 or 6q+3, or 6q+5, where q is some integer.
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[tex] \huge \fbox{{ \purple{Ⲫⳙⲉ⳽ⲧⲓⲟⲛ :-}}}[/tex]
Show that any positive odd integer is of the form, 6q+1 or 6q+3, or 6q+5, where q is some integer.
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Clearly, 6q + 1, 6q + 3 and 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3 and 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form 6q + 1 or 6q + 3 or 6q + 5.
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[tex]According \: to \: Euclid’s \: \\ division \: lemma[/tex]
[tex]a = bq + r[/tex]
[tex]a = 6q + r \: \: \: \: \: \: \: \: \: (1) \\ \\ where, (0 ≤ r < 6) \\ \\ So \: r \: can \: be \: either \\ 0, 1, 2, 3, 4 \: and \: 5[/tex]
[tex]Case \: 1: [/tex]
[tex]If r = 1, then \: equation (1) [/tex]
[tex]It becomes \: a = 6q + 1[/tex]
[tex]The \: Above \: equation \: will \: be \\ always \: an \: odd \: integer[/tex]
[tex]Case \: 2: [/tex]
[tex]If r = 3, then \: equation (1) [/tex]
[tex]It becomes \: a = 6q + 3 [/tex]
[tex]The \: Above \: equation \: will \: be \\ always \: an \: odd \: integer[/tex]
[tex]Case \: 3: [/tex]
[tex]If r = 5, then \: equation (1)[/tex]
[tex]It becomes \: a = 6q + 5 [/tex]
[tex]The \: Above \: equation \: will \: be \\ always \: an \: odd \: integer[/tex]
[tex]∴ Any \: odd \: integer \: is \: of \\ the \: form \: 6q + 1 \: or \\ 6q + 3 \: or \: 6q + 5 \\ \\ Hence \: proved[/tex]