- #1

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That is to say, if [tex]x, \alpha \in \mathbb{R}[/tex], then [tex]x=k \alpha + \delta[/tex] for some [tex]k \in \mathbb{Z}, [/tex] [tex]\delta \in \mathbb{R}[/tex] with [tex]0 \leq \delta < \alpha[/tex] where [tex]k, \delta[/tex] are unique.

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- Thread starter Doom of Doom
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- #1

- 86

- 0

That is to say, if [tex]x, \alpha \in \mathbb{R}[/tex], then [tex]x=k \alpha + \delta[/tex] for some [tex]k \in \mathbb{Z}, [/tex] [tex]\delta \in \mathbb{R}[/tex] with [tex]0 \leq \delta < \alpha[/tex] where [tex]k, \delta[/tex] are unique.

- #2

mathman

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- #3

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Yeah, but how can I show existence?

- #4

mathman

Science Advisor

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Yeah, but how can I show existence?

The sequence na for n=1,2,... is unbounded. Therefore for some n, na>x. Find lowest bound, subtract 1 and you will have k. ka<=x, (k+1)a>x, so x-ka(remainder)<a

- #5

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or even more straightforward, let k = floor(x/a)

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