# Count Primary C1

COUNTING1
We look at ways to count MANY.
MANY exists in time as repetition, and in space as multiplicity.
Both can be represented as strokes.
MANY strokes can be rearranged in icons so that there are four strokes in the icon 4 etc.
Then a total T can be counted in e.g. 4s by repeating the process ‘from T take away 4’.
This can be iconised as ‘T-4’.
The repeated process ‘form T take away 4s’ can be iconised as ‘T/4’
This makes it possible to predict the counting result through a calculation using the ‘recount-equation’ T = (T/b)*b, where the number T/b is called a per-number describing the total and the bundle-size.
Thus counting a total of 8 in 2s there are T/b = 8/2 = 4 = 4 per 1.
Changing units is another example of a recounting first telling how a given total can be counted into different units e.g.
T = 4\$ = 5kg producing a per-number 4\$/5kg.
Thus to answer the question ‘ 7kg=?\$’ we just have to recount the 7 in 5s: T = 7kg = (7/5)*5kg = (7/5)*4\$ = 5 3/5\$.
In most cultures ten is chosen as a standard bundle-size thus bundling and stacking in bundles of tens.
In this way a total T becomes a many-stack, a polynomial, consisting of a number of unbundled, a number of bundled, a number of bundles of bundled, etc.

Q: How to count MANY?
A: By bundling and stacking the total T predicted by T = (T/b)*b

Q: How to recount 8 in 3s: T= 8 = ? 3s
A: T= 8= ?*3= ?3s, T= 8=(8/3)*3 = 2*3 +2 = 2*3 +2/3*3 = 2 2/3*3

Q: How to recount 6kg in \$: T=6kg=?\$
A: If 4kg = 2\$ then 6kg = (6/4)*4kg = (6/4)*2\$ = 3\$

Q: How to count in standard bundles?
A: Bundling bundles gives a multiple stack, a stock or polynomial:
T = 423 = 4BundleBundle + 2Bundle + 3 = 4tenten2ten3 = 4*B^2 + 2*B + 3