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Do you remember when you first learned how to add? Remember how your teachers told you 1 + 1 = 2 ???? well thats not always true.... 1+1=1 is a true statement. 1+1=10 is also a true statement. It all depends on the basis and context... another thought what does the + mean?
Case 1: 1+1 = 1
Take the first example 1+1=1. Well if it is a severely limited environment (from our perspective, as the object existing in that environment may very well be happy with its capabilities and existance) where for example we only have one bit to represent our math and that bit is a binary representation, so only 0 and 1 are valid numbers in our system, well 1+1 represents an overflow. The answer could be 1, but it could also be 0. Its pretty undefined to say it will always be 1, but then again, it may be always 1, it just depends on how it treats an overflow. And what of the '+'? What does the plus sign represent? How does it act? In binary math (everything is base 2):
But notice the answer. In our world (the base 10 world) where we can map into base 2 quite easily and see potential carry digits the answer appears obvious.
In this case however, the entity stuck in a one bit binary representation would see this:
0 ?
The entity may see 0 or the entity may see 1. It all depends on how overflow is handled. So here is a case where 1 + 1 isn't 2!!

Case 2: 1+1 = 10

This argument is much simpler. In the base 2 world we have this:


Case 3: 1+1 = 5

When you say would 1 plus 1 be 5? Well say perhaps we gave another condition: Everything to the left and to the right of an equals sign represented a collection of numbers, and the equals sign shows membership. If we had that condition we would be saying redundantly, 1 and 1 are members of a set. 5 is also a member of a set. And both sets contain both at least one '1' and at least one '5'.

Interesting note to ponder.....

Can you think of a way to quickly calculate the square of a number in your head? Say someone said to you give me the square of 75. That's 75*75.... Got it yet? It turns out to be quite easy... In high school, I stumbled upon a quick formula and I used to earn extra lunch money by beating people on their TI-30 calculators (I used my brain). The answer is below but please read on for now....

Side Note

So now you may be wondering, what is the point here? Well this is an introduction to long integer arithmetic. When using computers and long digit math, the machine if left to itself will introduce round off errors when its bitsize is exceeded. For example, if a computer could only deal with numbers less than 2 and greater than -2, well we would introduce round off error potential once we exceed the limits of our computer. In the broader case, will the computer deliver the exact answer of two numbers multiplied together with the first number consisting of 3593 digits and the second number having 4217 digits? Your desktop calculator will display an error or overflow or something very big as a decimal number. Perhaps it might say something like 9.9X10**99 but that is not the correct answer (though from the calculator's perspective, that may very well be the correct answer).

I stumbled on a solution to long digit math that was easy to implement on a computer back in 1983 when I was a math major at CSU, Chico... After developing some theories related to it, I started asking around to see if anyone had knowledge of any of these theories... After a long search, I found out about Trachtenberg.... You can order the book here

As a POW in the World War, he came up with similar theories in math as a way to fight his mind from going insane from boredom. My theories turned out to be a specialized extension of his but I didn't pursue it any farther at that point because his were in themselves quite interesting. In 1985, I received my Computer Science degree from CSU, Chico.

Back to Weird Math...

The case of 75*75 is solvable using his methodology. Its quite simple:

1) 5*5 is 25 (save those)
2) 5*7 = 35 Now take 35*2 = 70 (save that)
3) 7*7 = 49 (save this)

Now arrange it all:

5625 <-- Thats the answer!!!!!

Note we never had a carry greater than 2.... That in itself is quite interesting. You can extend this
formula to any number of digits.

To solve this for 2 digits the actual formula is this: Picture 75 as being represented by 7*10+5*1.

This abstracts to a*10 + b*1 ==> So (10a+b)**2 ==> 100a**2 + 2*10ab + b**2
==> 100aa + 20ab + bb

so plugging in 75*75 ==> 100(7)(7) + 20(7)(5) + 5(5) ==> 4900 + 700 + 25 ==> 5625

You can derive the formula for 3 digits, and keep working until you have your n*m formula...

It turns out with numbers ending in 5 there is an even simpler solution:

(n)* (n+1) + 25

Here n = 7

so its 7 * (7+1) + 25 ==> 7*8 + 25 --> 56 + 25 ==> 5625

Note the + isn't the + you are probably thinking of. Its closer to an OR but not quite. If you were to picture the 56 being in one word (of a computer address space) and the 25 being in an adjacent word. Putting the two words together would read 5625.....

Another interesting point is noting the '1'. Where did the 1 come from, and what does it signify? Its interesting to note this happens to be our carry digit... but why is it here? What does it mean to fold it into the left side of the equation? Is there a general scheme this can fit into?

This is the beginning of the theory for handling long number arithmetic with no roundoff errors.

Anytime the formula asks you to save a quantity (like 5*5) its like putting it off in a 'word' of the computer....

The 2x2 Pattern | The 2x3 Pattern

The 3x2 Pattern | The 3x3 Pattern | The 4x4 Pattern

Odd digit theory (New stuff)

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Get the Trachtenberg Book! (Speed Math)

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