Chapter 1: Real Numbers

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Real Numbers Solutions


Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.


Suppose a be any positive integer and b = 6. Then, by Euclid’s algorithm, ๐’‚ = ๐Ÿ”๐’’ + ๐’“, for some integer ๐’’ ≥ ๐ŸŽ, and ๐’“ = ๐ŸŽ,๐Ÿ,๐Ÿ,๐Ÿ‘,๐Ÿ’,๐Ÿ“, because of ๐ŸŽ≤๐’“<๐Ÿ”.

Now substituting the value of r, we get,

If r = 0, then ๐’‚ = ๐Ÿ”๐’’

Similarly, for ๐’“= ๐Ÿ,๐Ÿ,๐Ÿ‘,๐Ÿ’ and 5, the value of a is ๐Ÿ”๐’’+๐Ÿ,๐Ÿ”๐’’+๐Ÿ,๐Ÿ”๐’’+๐Ÿ‘,๐Ÿ”๐’’+๐Ÿ’ and ๐Ÿ”๐’’+๐Ÿ“, respectively.

If ๐’‚ = ๐Ÿ”๐’’,๐’‚=๐Ÿ”๐’’+๐Ÿ ๐š๐ง๐ ๐’‚ = ๐Ÿ”๐’’+๐Ÿ’, then a is an even number and divisible by 2.

A positive integer can be either even or odd. Therefore, any positive odd integer is of the form of ๐Ÿ”๐’’+๐Ÿ,๐Ÿ”๐’’+๐Ÿ‘ ๐š๐ง๐ ๐Ÿ”๐’’+๐Ÿ“, where q is some integer.


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