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== Introduction ==
Other than counting, addition is perhaps the most basic mathematical operation. As such there are relatively few techniques to improve it's efficiency, those techniques that exist are basically ways to organise the calculation to make it simpler to compute mentally. Subtraction is similar, but in this case there are also some techniques to help with the carry/borrow process which unlike addition can run across multiple digits. You will no doubt already use most of the addition and subtraction techniques described here intuitively without thinking about them, many of the techniques may seem to be so basic that they don't deserve explanation, but it is worth re-iterating them even if just to make you aware that you are probably already using techniques that can be expanded upon to improve your arithmetic ability.<br />
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== Hukommelse mot kalkulering==

Det er et grunnleggende faktum at jo mer du husker, jo mindre trenger du å regne.<br/>
Det første du lærte om tall, var hvordan å telle. Du kan tenke på det som evnen til å legge én til hvilket som helst tall, f. eks. 0+1=1, 4+1=5, 7+1=8, 9+1=10, 10+1=11, 19+1=20, 56+1=57. Du kan instinktivt svaret på disse stykkene, fordi du har øvd på å huske dem siden ung alder (telle opp til 100).<br />
Når du har lært å telle, lærer du å legge sammen tall ved telling, å legge til en om gangen (først brukte du kanskje fingrene), helt til du fant svaret. Addering blir definert som å stadig legge til én, f. eks.

*'''6 + 3'''
  <math>
    \begin{align}
    6+3 & = 6+1+1+1 \\
        & = 7+1+1 \\
        & = 8+1 \\
        & =9
    \end{align}
  </math>
Det tat ikke lang tid før du husker summene av små tall, så du slipper å telle for å finne svaret på små stykker.<br />
Dette mønstret gjentar seg om og om igjen. Når du lærer en ny algoritme, vil du automatisk, etter en stund, lære svar (ofte små svar) så du slipper å gå igjennom hele algoritmen for å få svaret. Dette er kanskje den første og mest grunnleggende aritmetiske teknikken, '''husk, så slipper du å regne'''.<br />
Det er selvsagt umulig å huske alle svarene på alle algoritmene, men det er sikkert at å huske en del er 

Now it is obviously impossible to memorise the results of all possible calculations, but it is clear that a certain amount of memorisation is essential, and that some additional memorisation above the minimum requirement will vastly improve the efficiency of many calculations. Where to draw the line is difficult to say, but the following would be good basic set of results to memorise, (most of us will have already memorised these results long ago).

*'''Alle summer opp til 10.'''
  <math>
    \begin{align}
      2+6&=? \\
      5+4&=? \\
      3+4&=? \\
      6+4&=? \\
      etc.
    \end{align}
  </math>
*'''All compositions of 10'''
  <math>
    \begin{align}
      1+?&=10 \\
      2+?&=10 \\
      3+?&=10 \\
      4+?&=10 \\
      etc.
    \end{align}
  </math>

Just memorising the above lets you calculate any sum of two single digit numbers by partitioning and reorganising (see later), but it saves a lot of time to also memorise the sum of all pairs of single digit numbers and not just those sums up to 10. e.g.
*'''All single digit sums'''
  <math>
    \begin{align}
      2+9&=? \\
      8+8&=? \\
      4+7&=? \\
      7+6&=? \\
      9+7&=? \\
      etc.
    \end{align}
  </math>
Note that is is perhaps surprising how few people have memorised the above sums. You may think you have them memorised, but consider carefully; are you actually recovering the results from memory or are you working them out by partitioning and reorganisation. (e.g. Do you know that 7+6=13, or do you think "it's one less than 7+7", or "it's one more than 6+6". Do you actually know that 9+7=16, or are you thinking "9+7 is the same as 10+6".<br />
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== Addition is Commutative ==

One of the fist things you learn about addition is that it is commutative. i.e. it doesn't matter whether you add 5 to 6 or add 6 to 5, the answer is the same no matter what order the terms are in. e.g.:
  <math>
    \begin{align}
      2+9   &= 9+2   &= 11 \\
      4+15  &= 15+4  &=19\\
      7+53  &= 53+7  &=60\\
      23+61 &= 61+23 &=84 \\
      etc. \\
      generally: \\
      a+b &= b+a
    \end{align}
  </math>

This idea is implicitly understood by everyone, (e.g. if asked to do 3+54 you would almost certainly think of 54+3 before coming up with 57), but it is important to explicitly state that '''addition is commutative''' as this rule is important in many addition algorithms. (It should also be noted that other mathematical operators are not commutative, e.g. 6-5 is '''''not''''' the same as 5-6.)<br />
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== Addition is Associative ==

Another of the most basic addition laws is that addition is associative. i.e. if you have a number of additions to perform it doesn't matter which order you do them in. e.g.:

  <math>
    \begin{align}
      (2+9)+5  &= 2+(9+5)  &= 16 \\
      (4+15)+7 &= 4+(15+7) &=19\\
      etc. \\
      generally: \\
      (a+b)+c &= a+(b+c)
    \end{align}
  </math>



==Partitioning and Reorganising==

== The First Technique: Extended Counting ==

Soon after you have memorised how to count you then quickly learn how to add one to any number, e.g. you know that 7999+1=8000 even though you have almost certainly never counted from 1 to 8000! Note that you do this calculation without formally adding the digits and carrying the result. e.g. you don't do this:

  <math>
    \begin{matrix}
      8 & 9 & 9 & 9 & \\
        &   & _1& 1 & +1 \\
      \hline & & & 0
    \end{matrix}
      \quad \Rightarrow \quad
    \begin{matrix}
      8 & 9 & 9 & 9 & \\
        & _1&   & 1 & +1 \\
      \hline & & 0 & 0
    \end{matrix}
      \quad \Rightarrow \quad
    \begin{matrix}
      8 & 9 & 9 & 9 & \\
      _1&   &   & 1 & +1 \\
      \hline & 0 & 0 & 0
    \end{matrix}
      \quad \Rightarrow \quad
    \begin{matrix}
      8 & 9 & 9 & 9 & \\
        &   &   & 1 & +1 \\
      \hline 9 & 0 & 0 & 0
    \end{matrix}
  </math>

You instead instinctively know that when adding one to any number there are only two possible outcomes.

# If the last digit is not a nine then the answer is just the same number with the last digit increased by one
# If the last digits are nines then you replace the nines with zeroes and add 1 to the first digit (working from right to left) that isn't a nine.



== Subtracting 9 ==

== Subtraction from a power of 10 ==

You will see in the multiplication section the Vedic Mathematics Sutra '''Vertically and Crosswise''' is used to multiply numbers near a power of 10, (e.g. 10, 100, 1000, etc.). The first step in this technique is to subtract the numbers you are working with from the nearest power of 10. Luckily another sutra can help with this initial subtraction.<br />
'''All From 9 And The Last From 10''' tells us us how to subtract a number from the next highest power of 10; we simply subtract each digit of the number in question from 9 apart from the last one which we subtract from 10. It's as simple as that. e.g.<br />
<br />
*'''Subtract 8675 from 10000'''
 <math>
  \begin{matrix}
    9 & 9 & 9 & 10 \\
    8 & 6 & 7 & 5 & - \\
    \downarrow & \downarrow & \downarrow & \downarrow \\
    1 & 3 & 2 & 5
  \end{matrix}
 </math>
'''10000-8675=1325'''<br />
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The number you are subtracting must have the same number of digits as the number of zeroes in the power of 10 you are subtracting from. If your number has less digits than this you must pad the number with leading zeroes. e.g.
*'''Subtract 875 from 100000'''
 <math>
  \begin{matrix}
    9 & 9 & 9 & 9 & 10 \\
    0 & 0 & 8 & 7 & 5 & - \\
    \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\
    9 & 9 & 1 & 2 & 5
  \end{matrix}
 </math>
'''100000-875=99125'''<br />
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