# B Missing and Extraneous Solutions

#### FAS1998

Early in algebra I was taught that you could perform any action on both sides of an equation. Later I was taught that you can't perform all actions on both sides of an equation without the possibility of gaining/losing solutions.

What are the actual rules for what can be done to both sides of an equation without gaining/losing solutions.

I understand that if we have an equation like x^2 = x, and we divide by x, we get x = 1, losing the solution x = 0. In general dividing by x can be problematic because the division is not defined at 0.

But are there more general rules for what is and isn't allowed though? The Wikipedia page mentions something about operations being invertible, but doesn't go into much detail.

#### fresh_42

Mentor
2018 Award
For $A=B \Longrightarrow C=D$ you can perform any operation on $A=B$ as long as you do the same on both sides. The difficulty is on the way back. If we have a calculation like $A=B \Longrightarrow \ldots \Longrightarrow C=D$ then we have only derived a necessary condition, something which has to be true. It does not mean, that it is sufficient, as your example with $x=x^2$ shows.

Thus it is the way back that makes the problems. There is no way back from $x=1$ to $x=0$. I find it easiest to avoid all operations which do not allow a reverse. My solution for your example is $x=x^2 \Longrightarrow x-x^2=0 \Longrightarrow x\cdot (x-1) = 0 \Longrightarrow x= 0 \text{ or } x=1$ and all implications remain reversible.

#### FAS1998

For $A=B \Longrightarrow C=D$ you can perform any operation on $A=B$ as long as you do the same on both sides. The difficulty is on the way back. If we have a calculation like $A=B \Longrightarrow \ldots \Longrightarrow C=D$ then we have only derived a necessary condition, something which has to be true. It does not mean, that it is sufficient, as your example with $x=x^2$ shows.

Thus it is the way back that makes the problems. There is no way back from $x=1$ to $x=0$. I find it easiest to avoid all operations which do not allow a reverse. My solution for your example is $x=x^2 \Longrightarrow x-x^2=0 \Longrightarrow x\cdot (x-1) = 0 \Longrightarrow x= 0 \text{ or } x=1$ and all implications remain reversible.
How do we know if an operation is reversible? I believe addition and subtraction are reversible, but I'm not sure why exactly.

#### symbolipoint

Homework Helper
Gold Member
How do we know if an operation is reversible? I believe addition and subtraction are reversible, but I'm not sure why exactly.
Inverse Operations; Identity Elements.

#### symbolipoint

Homework Helper
Gold Member
FAS1998, regarding post #1,
Do you have a couple of specific examples?

#### fresh_42

Mentor
2018 Award
How do we know if an operation is reversible? I believe addition and subtraction are reversible, but I'm not sure why exactly.
The formal answer would be, because they are groups. This means addition can always be reversed by subtraction. Multiplication as well, by division. The only fact to consider is, that $0$ does not belong to the multiplicative group, so multiplying or dividing by zero is forbidden. And, yes, multiplying, too. At least in the context here. Although we may conclude $A=B \Longrightarrow 0=A\cdot 0 = B \cdot = 0$, we would lose any information, so it is useless to do.

#### jbriggs444

Homework Helper
The formal answer would be, because they are groups. This means addition can always be reversed by subtraction. Multiplication as well, by division. The only fact to consider is, that $0$ does not belong to the multiplicative group, so multiplying or dividing by zero is forbidden. And, yes, multiplying, too. At least in the context here. Although we may conclude $A=B \Longrightarrow 0=A\cdot 0 = B \cdot = 0$, we would lose any information, so it is useless to do.
If we are considering transformations in which both side of an equality are multiplied or divided by a fixed constant or have a fixed constant added or subtracted, this seems apt.

But what if we are faced with $x=y \Longrightarrow x^2 = y^2$?

This transformation is irreversible. And yet it involves multiplication of both sides by the same value. It would seem that there is more to it than just group membership. I find myself unable to articulate an answer.

#### fresh_42

Mentor
2018 Award
But what if we are faced with $x=y \Longrightarrow x^2 = y^2$?

This transformation is irreversible. And yet it involves multiplication of both sides by the same value. It would seem that there is more to it than just group membership. I find myself unable to articulate an answer.
Regard the details! I have said ...
For $A=B \Longrightarrow C=D$ you can perform any operation on $A=B$ as long as you do the same on both sides.
... and squaring is not the same on both sides.

Doing the same meant $x=y \Longrightarrow x^2 = xy \text{ or } xy=y^2$ in which case no problem will occur. You implicitly allowed $x=-y$ by writing $x^2 =y^2$. If we already have a situation $a=x^2=y^2=b$ then we can only conclude $x=\pm y$ which is reversible.

#### Mark44

Mentor
And yet it involves multiplication of both sides by the same value.
Not quite. The left side was multiplied by x while the right side was multiplied by y. You could make the case that x and y happen to be equal, but a better way to describe the operation is that you squared both sides to arrive at $x^2 = y^2$. Raising both sides to an even power is not reversible, but raising both sides to an odd power is reversible. For example $x = y \Leftrightarrow x^5 = x^5$ if we limit the discussion to real values only.

#### jbriggs444

Homework Helper
Not quite. The left side was multiplied by x while the right side was multiplied by y. You could make the case that x and y happen to be equal, but a better way to describe the operation is that you squared both sides to arrive at $x^2 = y^2$. Raising both sides to an even power is not reversible, but raising both sides to an odd power is reversible. For example $x = y \Leftrightarrow x^5 = x^5$ if we limit the discussion to real values only.
Yes. This is the sort of thing that I had in mind. Roughly speaking, one has two ways to justify a transformation from $x=y$ to $x^2=y^2$.

First way (as you've alluded):

One can multiply both sides by the same value. Whether that value happens to be x or happens to be y is irrelevant. The result is, for instance: $x^2 = xy$. Then one applies the rule of substitution. Since x=y we can replace the x on the right hand side by a y and arrive at $x^2=y^2$.

This is a reversible operation -- sort of. The substitution step is valid and reversible if you have a supporting assertion that x=y. Of course, when proceeding in reverse, it would be a bit silly to expect to have $x=y$ as a supporting assertion when the point of the exercise is to deduce $x=y$.

[One would also want to spend a few seconds justifying the reversibility of the multiply by x step to account for the possibility that x=0].

Second way (as you've also alluded):

Straight substitution. Given any function f(x), $x=y$ implies $f(x)=f(y)$. If we want this to be guaranteed reversible, we'd want the f(x) to be one to one and onto. $f(x)=x^2$ does not meet that requirement.

#### WWGD

Gold Member
It depends on the structures you are working with. Operations that preserve linear systems, polynomials , etc are different from each other.

"Missing and Extraneous Solutions"

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