If $x \equiv 1 \mod 3 $ then $x \equiv -2 \mod 6$?

Question

I have a feeling it should because it seems very close to if $$x \equiv 1 \mod 3 \ \ \text{and}\ \ x \equiv 1 \mod 6$$

since anything that is divisible by $6$ is also divisible by $3$.

But what is throwing me off here is the $-2$.

If I use the identities I get $x-1=3t$ for some integer $t$ and the other $x+2=6n$ for some integer $n$ and then I would get $x=3t+1$ and $x=6n-2$.

Answer 1

This is not true. $$7 \equiv 1 \pmod{3}$$

but

$$7 \not\equiv -2 \pmod{6}$$

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What you can say, however, is that if $x \equiv 1 \pmod{3}$, then exactly one of these two statements is true (remember that $-2$ is the same as $4$, modulo $6$).

$$x \equiv 1 \pmod{6}$$

$$x \equiv -2 \pmod{6}$$

Answer 2

Consider $7 \equiv 1 \mod 3$ and $7 \not\equiv -2 \mod 6$ for a counter example.

Show that $a \equiv b \mod m$ and $a \equiv b \mod n$ if, and only if, $a \equiv b \mod lcm (m,n)$.

Hint: (to the proof)

$(\implies)$ we have that $m | (a - b)$ and $n | (a -b)$, then by definiton of least commom multiple we have that $lcm (m,n) | (a - b)$.

Can you think the converse?

Answer 3

Because $6=2\times 3$ and $2,3$ are relatively prime:

$x\equiv -2 \pmod{6}$ if and only if

1. $x\equiv -2 \pmod{3}$, and
   
2. $x\equiv -2 \pmod{2}$.

The first condition is your hypothesis, but the second condition may or may not be true.

Answer 4

Hint $\ \ x \,=\, 1 + 3(\overbrace{\underbrace{r+2q}_{\large{r\, =\, 0,1}}}^{\large t}) \,\equiv\, 1,4\pmod 6.\,\ $ Also note $\ 4\equiv -2\pmod 6$

i.e. $\ 1+3\Bbb Z\, =\, 1 + 3\,(\{0,1\}\!+2\Bbb Z)\, =\, \{1,4\}\!+6\Bbb Z$

Answer 5

Since $x\equiv_3 1$ means $x=3k+1$, you have that

either $x=3(2h)+1=6h+1\equiv_6 1$

or $x=3(2h+1)+1=6h+4\equiv_6 4\equiv_6 -2$.