Simplifying (b-a)/(a-b): A Scientific Approach

[Nathi ORea]

Homework Statement

I know the answer is -1, but I don't get a step.

Relevant Equations

I have seen explanations online saying you can multiply the numerator by -1 to get; -1(a-b)/(a-b)
After then i do understand how (a-b) cancells out . I don't really get why you are allowed to use such trickery of multiplying just the numerator by -1
I think it has something to do with that -1/-1 equals 1, but don't get why you don't have to multiply the denominator as well.

Appreciate any help.

 

[Frabjous]

You are multiplying the numerator by -1 twice which is the same as multiplying it by 1 which you are always allowed to do.

 

[Nathi ORea]

You are multiplying the numerator by -1 twice which is the same as multiplying it by 1 which you are always allowed to do.

Thanks for the reply.
Doesn't multiplying the numerator twice equal 1 again?
-1 x -1 x (b-a)

 

[Frabjous]

Thanks for the reply.
Doesn't multiplying the numerator twice equal 1 again?
-1 x -1 x (b-a)

Yes. (b-a)=1x(b-a)=-1x-1x(b-a)=-1x(a-b)

 

[Frabjous]

You can also think about it as factoring -1 out of (b-a).

 

[DaveE]

Homework Statement: I know the answer is -1, but I don't get a step.
Relevant Equations: I have seen explanations online saying you can multiply the numerator by -1 to get; -1(a-b)/(a-b)
After then i do understand how (a-b) cancells out . I don't really get why you are allowed to use such trickery of multiplying just the numerator by -1
I think it has something to do with that -1/-1 equals 1, but don't get why you don't have to multiply the denominator as well.

Appreciate any help.

Yes. I think you understand this better than whoever you are reading. Just to repeat what the others said above:
$$ \frac {b-a}{a-b} = 1⋅\frac {b-a}{a-b} = \frac {-1}{-1}⋅\frac {b-a}{a-b} = (-1)⋅\frac {b-a}{-1⋅(a-b)} = (-1)⋅(\frac {b-a}{b-a}) = -1 $$
When you're not sure, do every step rigorously until it becomes obvious. Confusion either in explanations, or understanding, is often do to people jumping ahead and combining steps. Most of us end up skipping steps to do these problems once we are really familiar with the subject. Even in this example I've skipped some steps that I thought were too obvious to elaborate.

 

[hmmm27]

Or, you could multiply the denominator :

$$\frac{b-a}{a-b}=\frac{-1}{-1}\times \frac{b-a}{a-b} = \frac{-1}{1} \times \frac{b-a}{b-a} = -1$$

 

[Mark44]

You are multiplying the numerator by -1 twice which is the same as multiplying it by 1 which you are always allowed to do.

That's not what is happening. The original expression was ##\frac{b - a}{a - b}##.
To get to ##\frac{-1(a - b)}{a - b}##, there is no multiplication occurring -- in the numerator -1 is being pulled out as a factor, leaving (-b + a) as the other factor. Of course, -b + a is equal to a - b.

You can also think about it as factoring -1 out of (b-a).

Not "also" -- this is what is happening.

 

[PeroK]

Homework Statement: I know the answer is -1, but I don't get a step.
Relevant Equations: I have seen explanations online saying you can multiply the numerator by -1 to get; -1(a-b)/(a-b)
After then i do understand how (a-b) cancells out . I don't really get why you are allowed to use such trickery of multiplying just the numerator by -1
I think it has something to do with that -1/-1 equals 1, but don't get why you don't have to multiply the denominator as well.

Appreciate any help.

First:$$b-a = -(a -b)$$Hence, if ##a \ne b##, you can cancel the common factor ##a -b## and$$\frac{b-a}{a-b} = -1$$For example: ##a = 5, b = 2, a -b = 3, b - a = -3## and $$\frac{b-a}{a-b} =\frac{-3}{3} = -1$$

 

[WWGD]

Thanks for the reply.
Doesn't multiplying the numerator twice equal 1 again?
-1 x -1 x (b-a)

Please use parentheses here.

 

[Nathi ORea]

I appreciate all the replies here guys. Heaps of different ways to think about it! You have all been a huge help!