# need help understanding denominator/numerator selection/choice

#### rjs

##### New member
This is my first post, and I would like to take this opportunity to say, "hello," to everyone. Now, to the matter at hand:

I am studying the free notes/practice tests at a site associated with/hosted [and run] by lamar university: the link is as follows

http://tutorial.math.lamar.edu/

Now, there is a problem (number seven, integer/exponents section) whose answer is unclear to me: here is the link to the solution's page:

http://tutorial.math.lamar.edu/Solutions/Alg/IntegerExponents/Prob7.aspx

Can someone explain to me how variable "n" is chosen to be the denominator, and "m" the numerator?"

It is not immediately obvious to me, and the explanation would be greatly appreciated.

Also, if the decision is "up in the air", and either variable would suffice (as seems to be the case: at least, in my opinion), how likely is such a problem to be on standardized tests (for example, the "clep" and other "credit by examination" tests), and, furthermore, how is there any way to know you are getting the answer right if two answers are, technically, viable and correct?

Again, any and all help and suggestion is greatly appreciated.

Thank You

Sincerely,

rjs

#### Romsek

##### Junior Member
This is my first post, and I would like to take this opportunity to say, "hello," to everyone. Now, to the matter at hand:

I am studying the free notes/practice tests at a site associated with/hosted [and run] by lamar university: the link is as follows

http://tutorial.math.lamar.edu/

Now, there is a problem (number seven, integer/exponents section) whose answer is unclear to me: here is the link to the solution's page:

http://tutorial.math.lamar.edu/Solutions/Alg/IntegerExponents/Prob7.aspx

Can someone explain to me how variable "n" is chosen to be the denominator, and "m" the numerator?"

It is not immediately obvious to me, and the explanation would be greatly appreciated.

Also, if the decision is "up in the air", and either variable would suffice (as seems to be the case: at least, in my opinion), how likely is such a problem to be on standardized tests (for example, the "clep" and other "credit by examination" tests), and, furthermore, how is there any way to know you are getting the answer right if two answers are, technically, viable and correct?

Again, any and all help and suggestion is greatly appreciated.

Thank You

Sincerely,

rjs
I think all you really need to clear this up is the fact that

$$\displaystyle \forall m\neq 0,~\dfrac{1}{m^k} = m^{-k}$$

the rest is just a bit of arithmetic manipulation

#### Jomo

##### Elite Member
This is my first post, and I would like to take this opportunity to say, "hello," to everyone. Now, to the matter at hand:

I am studying the free notes/practice tests at a site associated with/hosted [and run] by lamar university: the link is as follows

http://tutorial.math.lamar.edu/

Now, there is a problem (number seven, integer/exponents section) whose answer is unclear to me: here is the link to the solution's page:

http://tutorial.math.lamar.edu/Solutions/Alg/IntegerExponents/Prob7.aspx

Can someone explain to me how variable "n" is chosen to be the denominator, and "m" the numerator?"

It is not immediately obvious to me, and the explanation would be greatly appreciated.

Also, if the decision is "up in the air", and either variable would suffice (as seems to be the case: at least, in my opinion), how likely is such a problem to be on standardized tests (for example, the "clep" and other "credit by examination" tests), and, furthermore, how is there any way to know you are getting the answer right if two answers are, technically, viable and correct?

Again, any and all help and suggestion is greatly appreciated.

Thank You

Sincerely,

rjs
Subtract the exponents for the base m, larger exponent - smaller exponent and put the result where the larger exponent is. For example, r-9/r-4= 1/r(-4)-(-9) = 1/r5

#### Otis

##### Senior Member
… answer is unclear to me …

… how variable "n" is chosen to be the denominator, and "m" the numerator" …
Hi! Welcome to the boards. :cool:

In a power, a negative sign in front of the exponent means the reciprocal of the power. For example:

$$\displaystyle 2^{3} = 2\cdot2\cdot2$$

$$\displaystyle 2^{-3} = \dfrac{1}{2\cdot2\cdot2}$$

In prealgebra, we learn how to simplify compound ratios (i.e., ratios which have fractions in their numerator and/or denominator). For example:

$$\displaystyle \dfrac{1}{\frac{1}{2}} = \dfrac{1}{1} \cdot \dfrac{2}{1} = 2$$

Here is the same example rewritten using exponential notation:

$$\displaystyle \dfrac{1}{2^{-1}} = \dfrac{2}{1}$$

Look. The base 2 moved from the denominator to the numerator.

When we remove the negative sign, we get the reciprocal of the power. In other words, removing a negative sign from an exponent shifts the power from the denominator to the numerator OR from the numerator to the denominator.

On that answer page, Paul's Notes instructs you to use this property of simplification, when removing negative signs from exponent. (The property is shown on this page.)

$$\displaystyle a^{ - n} = \dfrac{1}{a^n}$$

In your exercise, all of the powers have negative signs, so each power moves from numerator to denominator (or vice versa), after removing the negative signs. (We take the reciprocal of each power.)

$$\displaystyle \displaystyle \frac{{{m^{ - 2}}{n^{ - 10}}}}{{{m^{ - 7}}{n^{ - 3}}}} = \frac{{{m^7}{n^3}}}{{{m^2}{n^{10}}}}$$

The final simplification is canceling common factors. The two factors of m in the denominator cancel with two of the factors of m in the numerator, leaving five factors of m, in the numerator.

Likewise, canceling common factors of n leaves seven factors of n, in the denominator.

After you practice these simplications a lot, the process will become clear. If you'd like more practice or explanations, then you could google keywords negative exponents examples and watch some videos or read lessons at other sites.

#### Otis

##### Senior Member
Subtract the exponents …
There is no subtraction, in this exercise.

#### Jomo

##### Elite Member
There is no subtraction, in this exercise.
I am sorry but I disagree with you. This is a subtraction problem.

In earlier grades you are asked to compute 7-(-2) and -9 - (-11) and now in algebra you are asked to simplify (A7B-11)/(A-2B-9). To me both problems are the same.

Why convert all the exponents to positive values before simplifying?

Of course, as an instructor you MUST explain why subtractions works and repeat yourself until they understand why you subtract. After all, subtraction tell you how much bigger one number is than another (at least in absolute value) and this is exactly what is going on here with the cancellation.

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#### Otis

##### Senior Member
… This is a subtraction problem …
Okay. I didn't notice the combined properties that you used appearing in the materials presented in the op.

May I suggest (when working with a confused student) that you first show a specific property, before using it?

Also, you confused me, by mixing two different properties into one. I didn't realize that; to me, you were implying that -9 is a larger number than -4. I confess that I stopped paying attention, at that point.

It would have helped me, had you first shown the property that says a ratio of powers having the same base can be simplified by subtracting the exponent in the denominator from the exponent in the numerator.

$$\displaystyle \dfrac{b^{m}}{b^{n}} = b^{m - n}$$

… Why convert all the exponents to positive values before simplifying? …
It's just another valid way of thinking. Do you object, when your students use other methods?

Now that I understand what you were doing, I like those properties, too. #### Jomo

##### Elite Member
Okay. I didn't notice the combined properties that you used appearing in the materials presented in the op.

May I suggest (when working with a confused student) that you first show a specific property, before using it?

Also, you confused me, by mixing two different properties into one. I didn't realize that; to me, you were implying that -9 is a larger number than -4. I confess that I stopped paying attention, at that point.

It would have helped me, had you first shown the property that says a ratio of powers having the same base can be simplified by subtracting the exponent in the denominator from the exponent in the numerator.

$$\displaystyle \dfrac{b^{m}}{b^{n}} = b^{m - n}$$

It's just another valid way of thinking. Do you object, when your students use other methods?

Now that I understand what you were doing, I like those properties, too. I am glad that we agree with each other and I clearly noted your suggestions. Thanks.

#### Jomo

##### Elite Member
It's just another valid way of thinking. Do you object, when your students use other methods?
No, I do not object if my students use another method. I firmly believe that students should be given the chance to understand a problem their way. The problem is that in a classroom setting you do not always get a chance to show different ways of thinking about a given problem (and when you do some students complain). I do think that students should be taught the most efficient way to do a problem and in this particular case since you have to subtract in the end then why add another step to get all positive exponents before subtracting.
I also think that students need to learn how to subtract even if one or both numbers are negative.

#### JeffM

##### Elite Member
http://tutorial.math.lamar.edu/Solutions/Alg/IntegerExponents/Prob7.aspx

Can someone explain to me how variable "n" is chosen to be the denominator, and "m" the numerator?"

It is not immediately obvious to me, and the explanation would be greatly appreciated.

Also, if the decision is "up in the air", and either variable would suffice (as seems to be the case: at least, in my opinion), how likely is such a problem to be on standardized tests (for example, the "clep" and other "credit by examination" tests), and, furthermore, how is there any way to know you are getting the answer right if two answers are, technically, viable and correct?
To be honest, I am not sure I even understand your question. As the other answers have explained, there are various ways to attack this problem, but I am going to attack it using your text's method while using baby steps because I am not sure where you are getting lost. And yes, such problems may appear on a standardized test because the answers are not at all arbitrary.

General principal

$$\displaystyle a > 0 \implies a^{-b} ==\equiv \dfrac{1}{a^b} \equiv \left ( \dfrac{1}{a} \right )^b.$$

$$\displaystyle \therefore a > 0 \implies \dfrac{1}{a^{-b}} = \dfrac{1}{\dfrac{1}{a^b}} = \dfrac{1}{1} * \dfrac{a^b}{1} = a^b.$$

In short, $$\displaystyle a > 0 \implies a^{-b} = \dfrac{1}{a^b} \text { and } \dfrac{1}{a^{-b}} = a^b.$$

So let's follow step by step what is going on in your text:

$$\displaystyle \dfrac{m^{-2}n^{-10}}{m^{-7}n^{-3}} = m^{-2} * n^{-10} * \dfrac{1}{m^{-7}} * \dfrac{1}{n^{-3}}=$$

$$\displaystyle \dfrac{1}{m^2} * \dfrac{1}{n^{10}} * m^7 * n^3 =$$

$$\displaystyle \dfrac{m^7n^3}{m^2n^{10}} =$$

$$\displaystyle \dfrac{m^2 * m^5n^3}{m^2 * n^{10}} =$$

$$\displaystyle \dfrac{m^5n^3}{n^{10}} =$$

$$\displaystyle \dfrac{m^5 * n^3}{n^7 * n^3} =$$

$$\displaystyle \dfrac{m^5}{n^7}.$$

If you do not understand any line of that, please say where and why you got lost.

That is the method briefly indicated in your text and explained in more detail by Otis, but some steps were not shown because they were assumed to be obvious.

Jomo has indicated an alternative method to get to the exact same result. It has fewer steps, but it may be slightly more prone to careless error.

In any case there is very good reason why m and n ended up where they did.

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#### Jomo

##### Elite Member
This is my first post, and I would like to take this opportunity to say, "hello," to everyone. Now, to the matter at hand:

I am studying the free notes/practice tests at a site associated with/hosted [and run] by lamar university: the link is as follows

http://tutorial.math.lamar.edu/

Now, there is a problem (number seven, integer/exponents section) whose answer is unclear to me: here is the link to the solution's page:

http://tutorial.math.lamar.edu/Solutions/Alg/IntegerExponents/Prob7.aspx

Can someone explain to me how variable "n" is chosen to be the denominator, and "m" the numerator?"

It is not immediately obvious to me, and the explanation would be greatly appreciated.

Also, if the decision is "up in the air", and either variable would suffice (as seems to be the case: at least, in my opinion), how likely is such a problem to be on standardized tests (for example, the "clep" and other "credit by examination" tests), and, furthermore, how is there any way to know you are getting the answer right if two answers are, technically, viable and correct?

Again, any and all help and suggestion is greatly appreciated.

Thank You

Sincerely,

rjs
One more comment. You think that the location of m and n can be interchanged. Let's look at that more carefully. Say m = 1 and n= 2. Then m/n = 1/2 while n/m = 2. Clearly the results are not the same. You may argue that m and n each have powers associated with them. Please try to see if you get the same results with powers for m and n. That is say for example we have m^2/n^3. That will be 1^2/2^3 = 1/8, while n^3/m^2 will be the reciprocal of 1/8 which is 8. This is not to say that you can't find values for m and n where you get the same results. Understand that if you can find even one set of values for m and n where m^2/n^3$$\displaystyle \neq$$n^3/m^2 then in fact the two are NOT equal.

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#### HallsofIvy

##### Elite Member
No, I do not object if my students use another method. I firmly believe that students should be given the chance to understand a problem their way.
I remember a student once complaining bitterly that I had marked a test problem wrong because he had used a different method than the one I had gone over in class to solve it. I don't think I ever convinced him that I had marked it wrong because, what ever method he used, he had got the wrong answer! A surprising number of students seem to think that the "right answer" to an equation or problem is what ever you get by following a given method and that has nothing to do with the answer actually satisfying the equation or conditions of the problem.

The problem is that in a classroom setting you do not always get a chance to show different ways of thinking about a given problem (and when you do some students complain). I do think that students should be taught the most efficient way to do a problem and in this particular case since you have to subtract in the end then why add another step to get all positive exponents before subtracting.
I also think that students need to learn how to subtract even if one or both numbers are negative.