next up previous
Next: Exercise 1. Up: Quiz1 Previous: Notations.

Formulas.

Let m and n be two relatively prime elements of an Euclidean domain R. (You may think R = $ \mbox{${\mathbb Z}$}$ or R = $ \mbox{${\mathbb Q}$}$[x].) Let s, t $ \in$ R be such that s m + t n = 1. For every a, b $ \in$ R there exists c $ \in$ R such that

($\displaystyle \forall$x $\displaystyle \in$ R)     $\displaystyle \left\{\vphantom{ \begin{array}{l} x \equiv a \mod{\, m} \\  x \equiv b \mod{\, n} \\  \end{array} }\right.$$\displaystyle \begin{array}{l} x \equiv a \mod{\, m} \\  x \equiv b \mod{\, n} \\  \end{array}$  $\displaystyle \iff$  x $\displaystyle \equiv$ c mod m n (1)

where a convenient c is given by

c  =  a + (b - as m = b + (a - b)t n (2)

Therefore, for every a, b $ \in$ R the system of equations

$\displaystyle \left\{\vphantom{ \begin{array}{l} x \equiv a \mod{\, m} \\  x \equiv b \mod{\, n} \\  \end{array} }\right.$$\displaystyle \begin{array}{l} x \equiv a \mod{\, m} \\  x \equiv b \mod{\, n} \\  \end{array}$ (3)

has a solution. We give also

1 + $\displaystyle {\frac{{1}}{{4}}}$ + $\displaystyle {\frac{{1}}{{16}}}$ + ... + $\displaystyle {\frac{{1}}{{2^{2k}}}}$ < 1 + $\displaystyle {\frac{{1}}{{2}}}$ + $\displaystyle {\frac{{1}}{{4}}}$ + $\displaystyle {\frac{{1}}{{8}}}$ + $\displaystyle {\frac{{1}}{{16}}}$ + ... + $\displaystyle {\frac{{1}}{{2^{2k}}}}$ < 2 (4)


next up previous
Next: Exercise 1. Up: Quiz1 Previous: Notations.
Marc Moreno Maza
2006-01-09