Earlier today, I was explaining to friends how to use

**polish hand magic**(which I've referred to as "absurdist arithmetic" ever since multiplying 3 by 7) to multiply numbers smaller than five and greater than ten. One of them observed, "It would be an interesting exercise to prove this by induction for all natural numbers." That got me thinking; it obviously works for negative numbers, and there's no reason it shouldn't work for numbers other than integers. He replied, "I doubt whether one can have pi fingers."
You can totally have pi fingers. You just have to be willing to carry from the left as well as from the right. Keeps you agile. The problem is merely practical. Furthermore, there’s no reason to prove it by induction on natural numbers. It's easier to prove without induction for ALL the numbers.

1. Let ↑ = λ n.n-5 [This amounts to "number of fingers up" for any n whatsoever, no matter how many fingers you think you have.]

2. Let ↓ = λ m.10-m ["number of fingers down"]

3. Fix x and y, arbitrary numbers (don't care which sort).

4. ↓(x)*↓(y) = (10-x)*(10-y) [by def. of ↓]

5. (10-x)*(10-y)=100-10y-10x+xy [because algebra, or so I've heard]

6. ↑(x)+↑(y) = (x-5)+(y-5) [by def. of ↑]

7. 10 * ((x-5)+(y-5)) = 10x+10y-100 [also because algebra]

8. Thus, 10 * (↑(x)+↑(y)) + (↓(x)*↓(y)) = xy [bippity boppity boo]

9. Therefore, for all z and w, 10(sum of fingers up) plus (product of fingers down) equals the product of z and w. [existential generalization]

Pi times 8, by hand:

Left hand: -(5-pi)+3=pi-2=1.14ish. Put this in the 10's place.

Right hand: (10-pi) * 2 =13.72ish.

Left hand Right hand

1.14 13.72

Cary ten over to the left.

2.14 3.72

Carry 0.14 over to the right.

2 3.72+1.4

2 5.2

Pi*8 is 25.2ish.

Wonder what it's like to have imaginary imaginary fingers. i*8???