- #1
Edgardo
- 706
- 17
I have a question about congruences involving fractions.
For integers a and b the following is defined:
a and b are congruent modulo m (m is a natural number) if there exists
an integer k such that k*m = a-b
[itex]a \equiv b (\mbox{mod } m) \Longleftrightarrow \exists k \in \mathbb{Z} : km = a-b[/itex]
For example:
[itex]13 \equiv 4 (\mbox{mod } 9)[/itex] because [itex]1 \cdot 9 = 13-4[/itex]On the Wolfram mathworld website there are other examples in (8):
[itex]2 \cdot 4 \equiv 1 (\mbox{mod } 7)[/itex]
[itex]3 \cdot 3 \equiv 2 (\mbox{mod } 7)[/itex]
[itex]6 \cdot 6 \equiv 1 (\mbox{mod } 7)[/itex]
So far, so good.
But then in (9) they write:
[itex]\frac{1}{2} \equiv 4 (\mbox{mod } 7)[/itex]
[itex]\frac{1}{4} \equiv 2 (\mbox{mod } 7)[/itex]
[itex]\frac{2}{3} \equiv 3 (\mbox{mod } 7)[/itex]
[itex]\frac{1}{6} \equiv 6 (\mbox{mod } 7)[/itex]
which I don't understand.
At first I thought that for fractions a and b the definition is just extended:
[itex]a \equiv b (\mbox{mod } m) \Longleftrightarrow \exists k \in \mathbb{Z} : km = a-b[/itex]
with a and b fractions (instead of just integers).
But the definition of congruence for fractions must be different since
there is no [itex]k \in \mathbb{N}[/itex] such that
[itex]\frac{1}{2} - 4 = k \cdot 7:[/itex]
[itex]\frac{1}{2} - 4 = k \cdot 7[/itex]
[itex]\Rightarrow \frac{1}{2} - \frac{8}{2} = k \cdot 7[/itex]
[itex]\Rightarrow -\frac{7}{2} = k \cdot 7[/itex]
[itex]\Rightarrow k=-\frac{1}{2}[/itex]My questions:
a) How are congruences defined for fractions? And why is (9) correct?
b) Does (8) imply (9) ?
For integers a and b the following is defined:
a and b are congruent modulo m (m is a natural number) if there exists
an integer k such that k*m = a-b
[itex]a \equiv b (\mbox{mod } m) \Longleftrightarrow \exists k \in \mathbb{Z} : km = a-b[/itex]
For example:
[itex]13 \equiv 4 (\mbox{mod } 9)[/itex] because [itex]1 \cdot 9 = 13-4[/itex]On the Wolfram mathworld website there are other examples in (8):
[itex]2 \cdot 4 \equiv 1 (\mbox{mod } 7)[/itex]
[itex]3 \cdot 3 \equiv 2 (\mbox{mod } 7)[/itex]
[itex]6 \cdot 6 \equiv 1 (\mbox{mod } 7)[/itex]
So far, so good.
But then in (9) they write:
[itex]\frac{1}{2} \equiv 4 (\mbox{mod } 7)[/itex]
[itex]\frac{1}{4} \equiv 2 (\mbox{mod } 7)[/itex]
[itex]\frac{2}{3} \equiv 3 (\mbox{mod } 7)[/itex]
[itex]\frac{1}{6} \equiv 6 (\mbox{mod } 7)[/itex]
which I don't understand.
At first I thought that for fractions a and b the definition is just extended:
[itex]a \equiv b (\mbox{mod } m) \Longleftrightarrow \exists k \in \mathbb{Z} : km = a-b[/itex]
with a and b fractions (instead of just integers).
But the definition of congruence for fractions must be different since
there is no [itex]k \in \mathbb{N}[/itex] such that
[itex]\frac{1}{2} - 4 = k \cdot 7:[/itex]
[itex]\frac{1}{2} - 4 = k \cdot 7[/itex]
[itex]\Rightarrow \frac{1}{2} - \frac{8}{2} = k \cdot 7[/itex]
[itex]\Rightarrow -\frac{7}{2} = k \cdot 7[/itex]
[itex]\Rightarrow k=-\frac{1}{2}[/itex]My questions:
a) How are congruences defined for fractions? And why is (9) correct?
b) Does (8) imply (9) ?
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