System of congruences with no solution

[vincentvance]

Hello,

I am trying to prove that the system

x ≡ 5 (mod 6)
x ≡ 7 (mod 15)

has no solution. This example is done in my textbook and they say one should first assume that there is a solution x, and that when taking classes mod 6 the second equation implies that 2x ≡ 2 (mod 6). After this the example continues and I can follow the remaining steps, but this particular part I don't understand. What does it mean to "take classes mod 6" and why does the second equation imply that 2x ≡ 2 (mod 6)?

Any help is greatly appreciated.

 

[Opalg]

Hello,

I am trying to prove that the system

x ≡ 5 (mod 6)
x ≡ 7 (mod 15)

has no solution. This example is done in my textbook and they say one should first assume that there is a solution x, and that when taking classes mod 6 the second equation implies that 2x ≡ 2 (mod 6). After this the example continues and I can follow the remaining steps, but this particular part I don't understand. What does it mean to "take classes mod 6" and why does the second equation imply that 2x ≡ 2 (mod 6)?

Any help is greatly appreciated.

You might find it easiest to understand this by working mod 3, using the fact that 6 and 15 are both multiples of 3. In fact, if $x \equiv 5\pmod6$ then $x\equiv2\pmod3$. But if $x\equiv7\pmod{15}$ then $x\equiv1\pmod3.$

 

[vincentvance]

You might find it easiest to understand this by working mod 3, using the fact that 6 and 15 are both multiples of 3. In fact, if $x \equiv 5\pmod6$ then $x\equiv2\pmod3$. But if $x\equiv7\pmod{15}$ then $x\equiv1\pmod3.$

I feel like such an idiot, but I don't understand why for example if x ≡ 5 (mod 6) then x ≡ 2 (mod 3).

 

[Opalg]

I don't understand why for example if x ≡ 5 (mod 6) then x ≡ 2 (mod 3).

Think of it from the basic definition. If x ≡ 5 (mod 6), it means that $x$ is a multiple of 6, plus 5. Say $x = 6k+5$, for some integer $k$. You can write that as $x = 3(2k+1) + 2$. In other words, $x$ is a multiple of 3, plus 2. That is the same as saying x ≡ 2 (mod 3).

 

[vincentvance]

Think of it from the basic definition. If x ≡ 5 (mod 6), it means that $x$ is a multiple of 6, plus 5. Say $x = 6k+5$, for some integer $k$. You can write that as $x = 3(2k+1) + 2$. In other words, $x$ is a multiple of 3, plus 2. That is the same as saying x ≡ 2 (mod 3).

Ah. Of course, now it makes sense. Thank you so much!