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Why can 'per' units be expressed as ^-1?

Hi there,

Short version: Why can the units g/L also be written as g L-1?

Long version:

During my chemistry A-level this was explained to me really clearly but frankly I've forgotten now. Now I've almost finished my degree and I've asked lots of people and nobody seems to be able to explain it.

Basically some units such as mol/litre, g/litre, or mEq/L can also be expressed as mol L-1, g L-1, and mEq L-1 respectively.

While I fully understand what they mean, and it doesn't impact my ability to work and interpret results (I'm qualifying as a nurse soon), I just don't understand where the ^-1 comes from, mathematically. I'm hoping someone methematically minded can explain this to me.

Thank you!
(edited 7 years ago)
Reply 1
Original post by BananaWhale
Hi there,

During my chemistry A-level this was explained to me really clearly but frankly I've forgotten now. Now I've almost finished my degree and I've asked lots of people and nobody seems to be able to explain it.

Basically some units such as mol/litre, g/litre, or mEq/L can also be expressed as mol L^-1, g L^-1, and mEq L^-1 respectively.

While I fully understand what they mean, and it doesn't impact my ability to work and interpret results (I'm qualifying as a nurse soon), I just don't understand where the ^-1 comes from, mathematically. I'm hoping someone methematicaly minded can explain this to me.

Thank you!


It is a rate so you are dividing whatever comes before 'per' by whatever comes afterwards.
Reply 2
first x^-1 = 1/x, so something like density = mass / volume = mass * 1/volume = mass * (volume)^-1. Hence the units are Kg(m^3)^-1 = kgm^-3
Reply 3
Original post by kkboyk
It is a rate so you are dividing whatever comes before 'per' by whatever comes afterwards.


I understand that.

What I don't get is how to rearrange the divide to end up with a unit-1
(edited 7 years ago)
Reply 4
Original post by BananaWhale
...


It's a useful bit of notation that mathematicians have come up with, so a^(-1) means 1/a.

i.e: when you say gL^(-1), it's just a nicer way of writing g/L since mathematicians are lazy and don't enjoy having to write fractions which take up two lines, they prefer to just write L^(-1) and let it be known that that means 1/L, which extends to gL^(-1) meaning g * 1/L = g/L.
Reply 5
In maths, if a number has a power of minus one then it means divide by that number. For example

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So metres per second = m/s =
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(edited 7 years ago)
Reply 6
Original post by Zacken
It's a useful bit of notation that mathematicians have come up with, so a^(-1) means 1/a.

i.e: when you say gL^(-1), it's just a nicer way of writing g/L since mathematicians are lazy and don't enjoy having to write fractions which take up two lines, they prefer to just write L^(-1) and let it be known that that means 1/L, which extends to gL^(-1) meaning g * 1/L = g/L.


I appreciate you trying to help, thank you. It's the maths behind it which I'm trying to understand. Like I said, I understand what the unit means regardless of how it's written, but what I don't understand is how you can rearrange the formula to end up with unit-1.

It's not something I need to know, it's just frustrating me that I don't, when my chemistry teacher made it really clear a few years ago.
(edited 7 years ago)
Reply 7
Wikipedia goes into the derivation of this, go on the link below and scroll down to the "Negative exponents" section, hopefully it answers your question.

https://en.wikipedia.org/wiki/Exponentiation#History_of_the_notation
As has been pointed out, x^(-1) = 1/x. Take the example of g/L. This is the same as g*(1/L). We can write (1/L) as L^(-1). Therefore g/L = gL^(-1).
Reply 9
Original post by BananaWhale
I appreciate you trying to help, thank you. It's the maths behind it which I'm trying to understand. Like I said, I understand what the unit means regardless of how it's written, but what I don't understand is how you can rearrange the formula to end up with unit-1.

It's not something I need to know, it's just frustrating me that I don't, when my chemistry teacher made it really clear a few years ago.


I suppose you could consider

a * a^(-1) = a^1 * a^(-1) = a^(1 - 1) = a^0 = 1.

So you know know that a^(-1) * a = 1.

Divide both sides by a:

a^(-1) = 1/a.
gL^-1 is the same as g * L^-1

As people above have stated L^-1 is the same as 1/L.

When you multiply g/1 (which is just g) and 1/L you get the fraction g/L.
(Remember multiplying fractions: times bottoms together and then tops, e.g. 1/2 * 1/4 = 1/8)

Hope this helped. :smile:

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