Multiplying Gives the other number

A counter example: $(1,-1,-1)$

Hence, it is disproven.

Assume that the first number and the second number is $a,b$ respectively. Since the third number is the product of the previous two, we have the third number as $ab$ .

In addition, the third number multiplied by the first number must equal the second number, and the third number multiplied by the second number must equal the first number. Therefore, $ab\cdot a=b$ and $ab\cdot b=a$ .

Solving these two gives $b\cdot(a^2-1)=b\cdot(a+1)\cdot(a-1)=0$ and $a\cdot(b^2-1)=a\cdot(b+1)\cdot(b-1)=0$ .

Solving these, using the null factor law, demonstrates that solutions include $a=b=ab=1$ ; $a=b=-1,ab=1$ ; $a=1,b=ab=-1$ ; $a=ab=-1,b=1$ ; and $a=b=ab=0$ . Thus, $a=b=ab=1$ or $a=b=ab=0$ are not the only solutions.

We can readily find all of the solutions to this.

If one of the unknowns equals $0$ - without loss of generality let it be $a$ - then from the first equation $c = 0$ , and from the third equation $b = 0$ . This combination satisfies all three equations.

If none of them is $0$ , then we can divide the first equation by the second and get $\frac ac = \frac ca \\ \Rightarrow a^2 = c^2 \\ \Rightarrow |a| = |c|.$ By symmetry, $|a| = |b| = |c|$ . Taking the modulus of both sides of the first equation, this gives us $|ab| = |c| \\ \Rightarrow |a||b| = |c| \\ \Rightarrow |a|^2 = |a| \\ \Rightarrow |a| = 1.$ Of course, the last deduction is valid only because we have postulated that $a \neq 0$ . And by symmetry, $|b| = |c| = 1$ likewise.

It's easy to see that each equation is satisfied iff it contains an even number of $-1$ s. All three equations contain the same unknowns, so this is just an even number of $-1$ s among $a$ , $b$ and $c$ . Thus the possible solutions are: $a = b = c = 0 \\ a = b = c = 1 \\ a = 1, b = c = -1 \\ b = 1, a = c = -1 \\ c = 1, a = b = -1$

$1 \times (-1) = -1$

$(-1) \times (-1) = 1$

$(-1) \times 1 = -1$

Hence false.

Here's a really quick proof to find the counter-example in few steps. We want to find a solutions which, in particular, is not $(0,0,0)$ . This means that all three values are different from $0$ , since (WLOG) $a=0 \Rightarrow b=0 \Rightarrow c=0$ . Now that we know that, just divide all the equations by $a^2$ . This has for effect to obtain the much simpler problem (defining $x = \frac{b}{a}, y = \frac{c}{a}$ ) :

$x=y , xy=1$

$\Leftrightarrow x=y, x^2=1$

But this is problem has trivial solutions $x=y=1$ and $x=y=-1$ . This yields the $(1,1,1)$ solution and also the counterexample $(1,-1,-1)$ . Of course, permutating a,b and c we get the full set of solutions.

if we use 1 & -1.we can fill those conditions.

so,the numbers don't need to be 0s for all 1s.

by 3 equations we get a condition : a^2 = b^2 = c^2 ...which is valid for a=-1, b=-1 and c=1

It simply states that we are talking about numbers, but it's not specified what kind of numbers: can be real numbers, complex numbers or whatever else you want. So, if you know about quaternions , it's probably the first and the most obvious solution that comes into mind:

$ij=k$

$jk=i$

$ki=j$

After thinking a little, you can figure out some more solutions in real numbers, like combinations of number $1$ and two numbers $-1$ , but quaternions are probably the most classic solution to this one =)

a=(-1), b=1, c=(-1) is an example that also works. With a counterexample, we can conclude that the answer must be "no".

This is even solvable if one insists that $a\neq b\neq c$ as the basis of the quaternions satisfy all three equations. For the quaternions, we have three distinct square roots of -1, which we name $i,j,k$ such that $i^2= j^2 = k^2 = ijk = -1$ . It can be shown from this definition that $ij=k, jk=i,ki=j$ , thus if we set $a=i, b=j, c=k$ , we can have a solution to the problem in which not only are the three numbers distinct, but none of them are equal to 1, -1, or 0.

3 equations, 3 unknowns. I used the first equation to get the relation b=c/a. I substituted this relation into the third equation to get ca=c/a. solving for a=1 or -1. zero is not a solution, so the answer must be no.

Of course, the counterexample (1, -1, -1) works, but for those who couldn't just come up with it, here's how its done. Use the first and last equations to obtain that a=b/c=c/b (note that we can divide because we're ignoring the zero case). This tells us that a must then be 1 or -1 and the same holds true for b and c.

$\begin{aligned} ab &= c \\ bc &= a \\ ca &= b. \end{aligned}$ Multiplying equations above we have $\begin{aligned}(abc)^3 = abc\\ abc(abc+1)(abc-1)= 0\end{aligned}$ It is clear that either $abc =0$ , $abc =1$ or $abc =\boxed{-1}$ . Since the product of either $0$ or $1$ can never yield negative value. So it's false

$(-1, -1, 1), (-1, 1, -1), (0, 0, 0), (1, -1, -1), (1, 1, 1)$

The answer is no because they if these numbers are all 0s or 1s. It can't be because thr numbers are three different numbers: a,b and c.

An imaginary solution: (i, -i, 1)

It all depends on the triangle of multiplication of positives and negatives. With this, it becomes easy to visualize the negative cases.

One mistake that people might have done is to try the positive numbers instead of realizing that -1 and 1 are two different numbers. I was going to commit the same mistake but before saying yes, I realized that the numbers can also be negative. Thus, by trying out (-1, -1, 1), you will find that

1X(-1)=-1 (a=1, b=-1 and c=-1)

-1X(-1)=1 (b=-1, c=-1 and a=1)

1X(-1)=-1 (a=1, c=-1 and b=-1)

clearly not, but a^2=b^2=c^2 which 0 and 1 satisfies

Multiplicative inverse identity.

Let's all take a step back and think about the problem: "I have three numbers, and the product of any two of the numbers is equal to the other remaining number:" So, slow down and just look at the wording. "The Product of any two numbers(xy) is equal to(=) the other remaining number (z)" By definition no matter what single-digit, or multi-digit number you substitute for the variable, the statement is true. Therefore, it does, in fact, not need to be either a zero "0" or a one "1" in order for the statement to be true.

There is absolutely NO math involved in solving this particular problem. The only thing that you need is the ability to read without bias. Don't assume meanings based on context of socially asserted preferences. Read the statement as absolutely literal as you can.

Really, this is the very first step to any solution. Filter out bias, where possible.

ab = c _(1) ,bc = a _(2)

Dividing (1) by (2) we get a/c = c/a which is true for all 1s but not for 0s as 0/0 is undefined .