0

The problem is taken from my textbook in algebra and is given as:

Solve $x+4=3$ in $\mathbb{Z}_{7}$

From the euclidean algorithm I found that $2=4^{-1}\bmod(7)$

Now I'm not completly sure how to continue but I tried: $$x+4+4^{-1}=3+4^{-1} $$ $$x=3+4^{-1}=3+2=5 $$

However, the correct answer is $x=6$ and I'm not sure were I went wrong since the book don't have any examples so I would appreciate some help!

Bill Dubuque
  • 272,048
wroomwrooom
  • 77
  • 6
    $x+4\equiv 3\iff x\equiv -1$. And, of course, $-1\equiv 6$. All congruences being taken $\pmod 7$. – lulu Nov 30 '20 at 18:16
  • 14
    You're confusing the reciprocal and the negative. – Bernard Nov 30 '20 at 18:16
  • What book are you using? – Bill Dubuque Nov 30 '20 at 18:19
  • 3
    @wroom You are confusing the multiplicative inverse with the additive inverse – Ben Grossmann Nov 30 '20 at 18:32
  • Oh I see, haven't really gotten so far yet but will retry with additive inverse then! – wroomwrooom Nov 30 '20 at 18:36
  • @BillDubuque It's a swedish book but the name abstrakt algebra :) – wroomwrooom Nov 30 '20 at 18:37
  • 2
    Recall that $a + (-a) = 0$. Recall that $a\times (a^{-1}) = 1$. Don't confuse these. – JMoravitz Nov 30 '20 at 18:39
  • So, we have $x+4\equiv 3\pmod{7}$ implies that $x+4+(-4)\equiv 3+(-4)\pmod{7}$. Simplifying, we have $x\equiv -1\pmod{7}$ and if we wanted to, we can add $7$ since that is effectively the same as adding zero in the context of $\Bbb Z_7$ to get $x\equiv -1+7\equiv 6\pmod{7}$. This is all just a long-winded way of saying "you can subtract the same thing from both sides." – JMoravitz Nov 30 '20 at 18:41

1 Answers1

1

Solve $x+4=3$ in $\mathbb{Z}_{7}$

$(x+4)+3=3+3$

$x+(4+3)=6$

$x+0=6\;,\;$ indeed in $\;\mathbb{Z}_{7}\;,\;4+3=0\;.$

$x=6\;.$

Angelo
  • 12,328