Solving a⁻¹ - b⁻¹ = b - a /(ab)

[Nathi ORea]

TL;DR Summary: solving
a⁻¹ - b⁻¹ =
b - a /(ab)

I know that;

a⁻¹ - b⁻¹ =
b - a /(ab)

I am so lost as to how this answer is found. Can someone please give me a hint on any rules that allows you to solve this. A link perhaps

Thanks heaps.

 

[Infrared]

Note that $a^{-1}=\frac{1}{a}$ and $b^{-1}=\frac{1}{b}.$ If that's not clear, it would be a good idea to review how negative exponents work. We can put these fractions over the common denominator of $ab$ so:

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

 

[mcastillo356]

Note that $a^{-1}=\frac{1}{a}$ and $b^{-1}=\frac{1}{b}.$ If that's not clear, it would be a good idea to review how negative exponents work.

$$a^{-1}=a^{0-1}=\frac{a^0}{a^1}=\frac{1}{a}$$
When it comes to functions that are not constants, the rule changes:
Remark at "Calculus - A Complete Course 7th ed - Robert A. Adams and Christopher Essex":
"Do not confuse the $-1$ in $f^{-1}$ with an exponent. The inverse $f^{-1}$ is not the reciprocal $1/f$. If we want to denote the reciprocal $1/f(x)$ with an exponent we can write it as $\left(f(x)\right)^{-1}$

 

[BvU]

solving a⁻¹ - b⁻¹ = b - a /(ab)

There is no solving (in the sense of finding values, such that the equation is satisfied).
Do you mean "showing " ?

And: did you really mean a⁻¹ - b⁻¹ = b - a /(ab) ? Or did you mean: a⁻¹ - b⁻¹ = (b - a) / (ab) ?

$\$

 

[Mark44]

TL;DR Summary: solving
a⁻¹ - b⁻¹ =
b - a /(ab)

As already noted, this problem is not about solving an equation, but rather, proving that it is a true statement, at least for $a \ne 0$ and $b \ne 0$.

And: did you really mean a⁻¹ - b⁻¹ = b - a /(ab) ? Or did you mean: a⁻¹ - b⁻¹ = (b - a) / (ab) ?

It can only be the latter.

@Nathi ORea, as written the right side of the equation is b - a /(ab), which means $b - \frac a {ab}$. Although you used parentheses on the right side, you didn't use enough of them. If you want to write $\frac{b - a}{ab}$ without using LaTeX, write this as (b - a)/(ab), as @BvU did.

 

[Nathi ORea]

Note that $a^{-1}=\frac{1}{a}$ and $b^{-1}=\frac{1}{b}.$ If that's not clear, it would be a good idea to review how negative exponents work. We can put these fractions over the common denominator of $ab$ so:

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

Thanks heaps for the reply.

I did know the negative exponents. I would have got as far as 1/a- 1/b. I should have included that. I just wanted to get this question in quickly before work, so when I got home I had something to work off.. lol

I don't really get what you did after that sorry..

1/a- 1/b to a/(ab) - b/(ab).

I am not sure how you did that? Can you please explain.

I get the next step from:

a/(ab) - b/(ab) to (a-b)/(ab)

I'm fine with that at least.

I appreciate your help.

 

[Nathi ORea]

$$a^{-1}=a^{0-1}=\frac{a^0}{a^1}=\frac{1}{a}$$
When it comes to functions that are not constants, the rule changes:
Remark at "Calculus - A Complete Course 7th ed - Robert A. Adams and Christopher Essex":
"Do not confuse the $-1$ in $f^{-1}$ with an exponent. The inverse $f^{-1}$ is not the reciprocal $1/f$. If we want to denote the reciprocal $1/f(x)$ with an exponent we can write it as $\left(f(x)\right)^{-1}$

I appreciate your answer.

I am a pretty untalented 43 year old who is slowly (very slowly) making his way though a year 11 NSW textbook.. That sentence goes completely over my head.. lol.

I do understand that logical pathway though (if that is the correct usage at all), just the leap from:

I didn't quite get. How are they the same?

 

[Nathi ORea]

There is no solving (in the sense of finding values, such that the equation is satisfied).
Do you mean "showing " ?

And: did you really mean a⁻¹ - b⁻¹ = b - a /(ab) ? Or did you mean: a⁻¹ - b⁻¹ = (b - a) / (ab) ?

$\$

Sorry. I am not good with the lingo. I'll remember that.

Thanks.

 

[Nathi ORea]

As already noted, this problem is not about solving an equation, but rather, proving that it is a true statement, at least for $a \ne 0$ and $b \ne 0$.

It can only be the latter.

@Nathi ORea, as written the right side of the equation is b - a /(ab), which means $b - \frac a {ab}$. Although you used parentheses on the right side, you didn't use enough of them. If you want to write $\frac{b - a}{ab}$ without using LaTeX, write this as (b - a)/(ab), as @BvU did.

Yup. I stuffed up there ;)

 

[Infrared]

I don't really get what you did after that sorry..

1/a- 1/b to a/(ab) - b/(ab).

If you have a fraction, you can always multiply the numerator and denominator of the fraction by the same (nonzero) number, without changing the value of the fraction, for example, $\frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20}.$ This should make sense- if you have $5$ times as much money to distribute equally among $5$ times as many recipients, for example, then the amount of money each person receives is unchanged.

In our case, we can multiply the numerator and denominator of $\frac{1}{a}$ by $b$ to see that it is equal to $\frac{b}{ab}$ and similarly $\frac{1}{b}$ is equal to $\frac{a}{ab}$ (you wrote this backwards in the message I quoted).

 

[PeroK]

$$a^{-1}=a^{0-1}=\frac{a^0}{a^1}=\frac{1}{a}$$
When it comes to functions that are not constants, the rule changes:
Remark at "Calculus - A Complete Course 7th ed - Robert A. Adams and Christopher Essex":
"Do not confuse the $-1$ in $f^{-1}$ with an exponent. The inverse $f^{-1}$ is not the reciprocal $1/f$. If we want to denote the reciprocal $1/f(x)$ with an exponent we can write it as $\left(f(x)\right)^{-1}$

This is only hindering the OP, who's not at this level yet.

 

[PeroK]

I am a pretty untalented 43 year old who is slowly (very slowly) making his way though a year 11 NSW textbook..

This problem is an example of using what is called a common denominator. You should have heard about them. The post above by @Infrared shows how useful this is when you have specific fractions, like
$$\dfrac 1 2 - \dfrac 1 3 = \dfrac 3 6 - \dfrac 2 6 = \dfrac 1 6$$Or, when you have algebra to do:
$$\dfrac 1 a - \dfrac 1 b = \dfrac {b}{ab} - \dfrac {a}{ab} = \dfrac{b-a}{ab}$$In the first case, we used a common denominator of $6 = 2 \times 3$. In the second case, the common denominator was $ab$.

 

[scottdave]

Thanks heaps for the reply.

I did know the negative exponents. I would have got as far as 1/a- 1/b. I should have included that. I just wanted to get this question in quickly before work, so when I got home I had something to work off.. lol

I don't really get what you did after that sorry..

1/a- 1/b to a/(ab) - b/(ab).

I am not sure how you did that? Can you please explain.

I get the next step from:

a/(ab) - b/(ab) to (a-b)/(ab)

I'm fine with that at least.

I appreciate your help.

Would you agree that we can multiply by (ab)/(ab) {when a and b are not zero} without changing the value?

So multiply (1/a) by (ab)/(ab) can be arranged as [(ab) / a ] * [ 1 / (ab) ] = b * [ 1 / (ab) ] = b / (ab).

Do the same for the (1/b) term then combine.

 

[Nathi ORea]

This problem is an example of using what is called a common denominator. You should have heard about them. The post above by @Infrared shows how useful this is when you have specific fractions, like
$$\dfrac 1 2 - \dfrac 1 3 = \dfrac 3 6 - \dfrac 2 6 = \dfrac 1 6$$Or, when you have algebra to do:
$$\dfrac 1 a - \dfrac 1 b = \dfrac {b}{ab} - \dfrac {a}{ab} = \dfrac{b-a}{ab}$$In the first case, we used a common denominator of $6 = 2 \times 3$. In the second case, the common denominator was $ab$.

Thanks for that. I went back and learnt how to do it. I appreciate you gave it a name, because it makes it easier to search it up.