Maths Problem

[daniel.oleary]

I understand how to calculate problems that use negative indices, however we have been asked to find out why a negative indice causes your fraction to flip. I have spent ages trying to find the answer to this question but I can't. I am about to give up. So if someone can answer this for me I would be very happy.

 

[pka]

I understand how to calculate problems that use negative indices, however we have been asked to find out why a negative indice causes your fraction to flip. I have spent ages trying to find the answer to this question but I can't. I am about to give up. So if someone can answer this for me I would be very happy.

It is just a matter of notation.
If \(\displaystyle x\ne 0\) then it has a multiplicative inverse, That is there is some \(\displaystyle y\) such that \(\displaystyle x\cdot y=1\).

We already know that \(\displaystyle x\cdot\dfrac{1}{x}=1\) thus it must the case that \(\displaystyle y=\dfrac{1}{x}\).

Moreover, we agree that \(\displaystyle x^0=1\). So to make the laws of exponents consistent we need \(\displaystyle \dfrac{1}{x}=x^{-1}\)

So that means \(\displaystyle x\cdot{x^{-1}}=1\).

Now we have \(\displaystyle \dfrac{a}{b}\cdot\dfrac{b}{a}=1\) meaning that \(\displaystyle \dfrac{b}{a}=\left(\dfrac{b}{a}\right)^{-1}\).

 

[soroban]

Hello, daniel.oleary!

I understand how to calculate problems that use negative indices.
However, we have been asked to find out why a negative indice causes your fraction to flip.
I have spent ages trying to find the answer to this question but I can't.

We have a fraction to a negative power: .\(\displaystyle \left(\dfrac{a}{b}\right)^{-n}\)

Then we have: .\(\displaystyle \dfrac{a^{-n}}{b^{-n}} \;=\;\dfrac{b^n}{a^n} \;=\;\left(\dfrac{b}{a}\right)^n \)

 

[Mrspi]

It is just a matter of notation.
If \(\displaystyle x\ne 0\) then it has a multiplicative inverse, That is there is some \(\displaystyle y\) such that \(\displaystyle x\cdot y=1\).

We already know that \(\displaystyle x\cdot\dfrac{1}{x}=1\) thus it must the case that \(\displaystyle y=\dfrac{1}{x}\).

Moreover, we agree that \(\displaystyle x^0=0\). So to make the laws of exponents consistent we need \(\displaystyle \dfrac{1}{x}=x^{-1}\)

So that means \(\displaystyle x\cdot{x^{-1}}=1\).

Now we have \(\displaystyle \dfrac{a}{b}\cdot\dfrac{b}{a}=1\) meaning that \(\displaystyle \dfrac{b}{a}=\left(\dfrac{b}{a}\right)^{-1}\).

I'm SURE you meant "we agree that \(\displaystyle x^0=1\).

 

[pka]

Thanks