Algebraic Simplification of (2p^2q - 3pq^2 + q^3)/(pq^2 - p^2q - 2p^3)

Dr.Peterson said:
Those two expressions are not equivalent. But no one is saying they are!

I don't see that anyone has written the second expression you show here. Read carefully.

You said this:

Steven said this:

These are equivalent, because q-p = -(p-q).
Click to expand...
Settled!
 
I never felt it customary, however, I will myself use the fewest number of negative signs possible. I particularly abhor leading negatives.
 
MarkFL said:
I never felt it customary, however, I will myself use the fewest number of negative signs possible. I particularly abhor leading negatives.
Click to expand...
It just math grammar. I would never mark it wrong if someone didn't write their final answer with the fewest number of negative signs.
 
Dr.Peterson said:
I myself would probably write [math]- \frac{ q \left( p - q \right) }{ p \left( p + q \right)},[/math]
Click to expand...
That minus sign in the front can easily be lost. If I recall correctly, you made that same statement (That minus sign in the front can easily be lost) in the past. I never said the OP's work was wrong, in fact if I recall what I said about their work was it looks good to me.
It has always been customary to distribute a minus sign in front of a subtraction.
 
MarkFL said:
I never felt it customary, however, I will myself use the fewest number of negative signs possible. I particularly abhor leading negatives.
Click to expand...
Up till this very moment, I have not understood what you are trying to point out. When I understand, maybe I would be able to make some good point out of it.
 
Steven G said:
All we are saying is that if you distribute the negative sign you will get -p(p-q) = p(q-p)
Click to expand...
It is now I am just seeing it that
[math]- p \left( p- q \right)= - p \left( - q + p \right) = p \left( q - p \right)[/math]So this is what everybody was saying all this while?
 
chijioke said:
It is now I am just seeing it that
[math]- p \left( p- q \right)= - p \left( - q + p \right) = p \left( q - p \right)[/math]So this is what everybody was saying all this while?
Click to expand...
Yes. I guess we assumed you'd been taught such things.

But you didn't check it out when I said this?
Dr.Peterson said:
These are equivalent, because q-p = -(p-q).
Click to expand...
Therefore p(q-p) = -p(p-q).

You're expected either to try it and see, or at least to ask about what you don't understand.

I guess this illustrates why we need to give complete answers sometimes, and not a;ways try to draw them out of students who may just not have the required background,
 
Dr.Peterson said:
This is simply not true
Click to expand...
In defense of Steven, I note those instructors/authors who claim that it is. (We know the type: "My way or the highway".)

Hence, I wouldn't be surprised to learn that some people end up believing it's an actual math rule. (I was certainly penalized by various instructors early on for violating their "law".) ;)
[imath]\;[/imath]
 
Otis said:
In defense of Steven, I note those instructors/authors who claim that it is. (We know the type: "My way or the highway".)

Hence, I wouldn't be surprised to learn that some people end up believing it's an actual math rule. (I was certainly penalized by various instructors early on for violating their "law".) ;)
[imath]\;[/imath]
Click to expand...
You have won. No more argument.
 
People can convey their message in a language with grammatical mistakes and their message is perfectly understood. The same thing is with math. I would never penalized a student for poor math grammar, but I would point it out, as I did.
I was told that it is better to have as few negative signs as possible. When one or more mathematicians tell you something about math, you think it is true. To be honest, if it were important enough for them to tell me about signs, I will surely tell the same to my students.
 
You must log in or register to reply here.
Top