Why is the answer "No solution"?

[Math9999]

Homework Statement

Solve 2x=1 in ℤ₄.

Homework Equations

None.

The Attempt at a Solution

The answer is "No solution" but I don't know why. Because this is a modular problem, so 4+1=5 and 2x=5, x=5/2. But why the answer is no solution? Can anyone please explain?

 

[fresh_42]

$\mathbb{Z}_4=\{0,1,2,3\}$ so there are neither $4$ nor $5$ and certainly no $\frac{5}{2}$. Have you tried to draw the multiplication table?

 

[Math9999]

So any answer you get with fractions, that means no solution?

 

[fresh_42]

So any answer you get with fractions, that means no solution?

No. Formally a division is a multiplication with the inverse element. So whether a division is allowed or not, only depends on whether there is an inverse or not. An inverse element of, say $a$, is the element $x$ that satisfies $a\cdot x=1.$ Such an element is unique, so we write $x= a^{-1}$ or - not so good - $\frac{1}{a}$. The question behind the exercise is: Why doesn't $2 \in \mathbb{Z}_4$ have an inverse? E.g. $3 \in \mathbb{Z}_4$ has one, because $3 \cdot 3 \equiv 9 \equiv 1 \operatorname{mod} 4$.

 

[Math9999]

And what about problem like 6x=5 in ℤ₈?

 

[WWGD]

So any answer you get with fractions, that means no solution?

Well, no, they are asking for a solution in $\mathbb Z_4 =\{0,1,2,3\}$ with standard modular addition and multiplication and this does not include fractions.

 

[fresh_42]

And what about problem like 6x=5 in ℤ₈?

Consider more generally two cases.

$6\cdot 4 \equiv 0 \operatorname{mod} 8$. Now try to find out, whether such an element can have an inverse or not. What will happen if there is an element $a$ for which there are elements $x,y$ with $a\cdot x = 0$ and $a\cdot y = 1$. Can this happen?

Now even if $6 \in \mathbb{Z}_8$ had no inverse, there could still have been an element $x$ with $6\cdot x \equiv 5 \operatorname{mod} 8$. But we can solve it the same way, we would solve it with integers. If $6$ has no inverse, we will not be allowed to divide by $6$. But $5 \cdot 5 \equiv 25 \equiv 1 \operatorname{mod} 8$ so we can divide by $5$. The inverse element to $5$ is also $5$, so
$$5 \cdot 6 \cdot x \equiv 5 \cdot 5 \equiv 1 \operatorname{mod} 8
\;\;\text{ and } \;\;
5 \cdot 6 \cdot x \equiv 30\cdot x \equiv 6 \cdot x \operatorname{mod} 8
$$ Thus the equation $6\cdot x \equiv 5 \operatorname{mod} 8$ is solvabel if and only if $6 \in \mathbb{Z}_8$ has an inverse, which we dealt with in the first part.

The moment you forget about divisions and consider them as what they are, multiplications, you can calculate like you're used to. The only advice here is, that it is better to write a little more and do a little less in mind to avoid mistakes.

And of course we can always write down the multiplication table and see which multiples of $6$ lead to $5$, which is the boring way and not practicable in say $\mathbb{Z}_{186}$.

 

[WWGD]

And what about problem like 6x=5 in ℤ₈?

Think of $6x=2(3x)$ will be even , and an even number minus 8 will also be even... Think in terms of the gcd of some numbers.

 

[Math9999]

Got it. Thanks to everyone.