# Mathematic notation (doubtless a stupid question)

On Wednesday, September 11, 2019 at 9:06:13 PM UTC+1, David Paste wrote:

because it is *per* second - the inverse, ie
10m --------------- second * second
1/x == x^-1 in this notation.
Compare, for instance, measuring carpet, which is in meters^2, ie 'square meters', meters * meters, and not a minus in sight.
HTH J^n
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On Wednesday, 11 September 2019 21:20:44 UTC+1, jkn wrote:

it's metres ;-)

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On Wed, 11 Sep 2019 13:06:10 -0700 (PDT), David Paste

'per second per second' is not 'seconds squared', it's 'per second squared' (perhaps you meant that but omitted the 'per').
'one tenth' or 1/10 can be written 10^-1. Metres per second per second can be written m/sec/sec or m/sec^2 or m.s^-2, the -ve sign indicating 'per' or 'one over' or division.
--

Chris

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On Wed, 11 Sep 2019 13:06:10 -0700, David Paste wrote:

It's because it's /per/ second, indicating dividing. ms^2 would be metre- second-seconds (i.e. distance times time times time), while acceleration is ms^-2, metres per second per second, distance divided by time divided by time. Positive powers are multiplication, negative powers are division (or "anti-multiplication"). x^2 is x*x, x^-2 is (1/x)/x. 2^2 is 4, 2^-2 is 0.25.
If you multiply acceleration (ms^-2) by time (s, or s^1), you add the powers - and get speed in ms^(-2+1) or ms^-1. If you divide acceleration by time, you subtract the powers - and get jerk in ms^(-2-1) or ms^-3. The same applies starting with distance, in m, or ms^0. Anything to the power zero is 1, so m and ms^0 are the same thing. Divide by time and you get ms^(0-1), ms^-1.
Mike
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On 11/09/2019 21:06, David Paste wrote:

Its a bit like the way you can do a division by multiplying by the reciprocal of the devisor...
So m/s^2, can become ms^-2, so its including the "per" (i.e. the division operation) into the exponent.
--
Cheers,

John.
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While all (or perhaps most) of the answers posted are correct, I don’t think any quite get to the basic reason.
For that you need to look at the laws of indices.
4 = 2^2 2= 2^1 1 = 2^0 1/2 = 2 ^ -1 1/4 = 2^ -2 = 1/(2^2)
I’ve use 2 for simplicity but the rules apply for other numbers.
Hopefully you can see the pattern.
If we substitute m for 2 then, in particular in the last line we get:
m^-2 = 1/(m^2) which is also 1/m * 1/m
The laws of indices ‘pop up’ in a number of places and can be very useful. They are the basis of Logarithms, can be used to find HCF and LCMs, .......
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