Fun With Math-Superpowerition: Inventing a Super Power Operation -- Comparing The Math Operators II

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Comparing The Math Operators II

Let's take two numbers 2's and see how they compare with the five math operators: addition, multiplication, exponentiation, superpowerition and tetration.

First, let's define tetration. It's similar to superpowerition, but there are differences:

The notation for tetration is expressed as ⁿa or a ↑↑ n
You solve the problem from right-to-left instead of from left-to-right. For example, You solve ⁴2 or 2 ↑↑ 4 like this: 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 4) = 2 ^ 16 = 65,536.
The resulting answers to tetration zoom up even faster than superpowerition because of the reverse order of solving the problem using an odd right-to-left method for combined exponents.

We'll be using "^^" to indicate a tetration operation for the most part.

That done, here are the comparisions.

2 + 2 = 4

2 × 2 = 4

2 ^ 2, or 2² = 4

2 ✱ 2 = 2 ^ 2 = 4

2 ^^ 2 = 2 ^ 2 = 4

Is it a coincidence that using two 2's with any of the five math operators results in 4?

Now let's go a step further: Let's take the numbers 2 and 3 and see how they compare.

2 + 3 = 5

3 + 2 = 5

2 × 3 = 6

3 × 2 = 6

2 ^ 3, or 2³ = 8

3 ^ 2, or 3² = 9

2 ✱ 3 = (2 ^ 2) ^ 2 = 4 ^ 2 = 16

3 ✱ 2 = 3 ^ 3 = 27

2 ^^ 3 = 2 ^ (2 ^ 2) = 2 ^ 4 = 16

3 ^^ 2 = 3 ^ 3 = 27

You notice that the exponentiation and superpowerition operators are not communative. By switching around the two digits in those two operations, you get a different answer. Also notice that the superpowerition and tetration operations for 2 and 3 in this order is 16 for both instances? It's also the same for the operations for 3 and 2 in this order with the answer of 27. If there are just the number "2" or "3" in the second argument for either of the superpowerition and tetration operations, then the exponent is solved the same way in both cases.

Now let's go a step further: Let's take the numbers 2 and 4 and see how they compare.

2 + 4 = 6

4 + 2 = 6

2 × 4 = 8

4 × 2 = 8

2 ^ 4, or 2⁴ = 16

4 ^ 2, or 4² = 16

2 ✱ 4 = ((2 ^ 2) ^ 2) ^ 2 = (4 ^ 2) ^ 2 = 16 ^ 2 = 256

4 ✱ 2 = 4 ^ 4 = 256

2 ^^ 4 = 2 ^ (2 ^ (2 ^ 2)) = 2 ^ (2 ^ 4) = 2 ^ 16 = 65,536

4 ^^ 2 = 4 ^ 4 = 256

In this case with the 2 and 4 digits, that the exponentiation and superpowerition operators are communative. By switching around the two digits in those two operations, you get the same answer in this case. As you can tell, there is a number "4" in the second argument for the superpowerition and tetration operations, so the answers will differ due to the problem solving methods of each operation.

Now let's go a step further: Let's take the numbers 2 and 5 and see how they compare.

2 + 5 = 7

5 + 2 = 7

2 × 5 = 10

5 × 2 = 10

2 ^ 5, or 2⁵ = 32

5 ^ 2, or 5² = 25

2 ✱ 5 = (((2 ^ 2) ^ 2) ^ 2) ^ 2 = ((4 ^ 2) ^ 2) ^ 2 = (16 ^ 2) ^ 2 = 256 ^ 2 = 65,536

5 ✱ 2 = 5 ^ 5 = 3,125

2 ^^ 5 = 2 ^ (2 ^ (2 ^ (2 ^ 2))) = 2 ^ (2 ^ (2 ^ 4)) = 2 ^ (2 ^ 16) = 2 ^ 65,536 = 2.00353 × 10^19,728

5 ^^ 2 = 5 ^ 5 = 3,125

You notice that the exponentiation and superpowerition operators are not communative. By switching around the two digits in those two operations, you get a different answer. Also, if you look at the tetration operation of 2 ^^ 5, the answer zoomed so high it went out of the atmosphere! You can only imagine what the results would be with numbers of "3" and higher for tetration with results too high for your basic calculator to handle. You'll need a university mainframe computer to handle this tetration operation for numbers "3" and higher. For example, 10 ^^ 2 would result in the answer of 10 billion, an answer that will bust any cheap calculatior, while 10 ^^ 3 results in 10^{10,000,000,000}, far too many digits for your personal computer to handle. You can read about tetration, as well as higher hyperoperations such as pentation and hexation at another website.

You may have also noticed that in the case of powers of two, you have this pattern:

2 ✱ 2 = 2 ^ 2 = 4

2 ✱ 3 = (2 ^ 2) ^ 2 = 4 ^ 2 = 16

2 ✱ 4 = ((2 ^ 2) ^ 2) ^ 2 = (4 ^ 2) ^ 2 = 16 ^ 2 = 256

2 ✱ 5 = (((2 ^ 2) ^ 2) ^ 2) ^ 2 = ((4 ^ 2) ^ 2) ^ 2 = 16 ^ 2 ^ 2 = 256 ^ 2 = 65,536

And this result:

2 ✱ 2 = 2 ^ 2 = 2 ^ (2) = 2 ^ 2 = 4

2 ✱ 3 = (2 ^ 2) ^ 2 = 2 ^ (2 × 2) = 2 ^ 4 = 16

2 ✱ 4 = ((2 ^ 2) ^ 2) ^ 2 = 2 ^ (2 × 2 × 2) = 2 ^ 8 = 256

2 ✱ 5 = (((2 ^ 2) ^ 2) ^ 2) ^ 2 = 2 ^ (2 × 2 × 2 × 2) = 2 ^ 16 = 65,536

In cases where you have more than one exponent, you take the numbers that are bordering the second and later exponentiation signs, multiply them together into a product, then take the base to the power of the product. You get the same answer!

You can also summarize the superpowerition operation with this formula:

a ✱ b = a ^ (a ^ (b - 1)) or a^{(a^{(b - 1)})}

This works for numbers 2 and higher. I have yet to test them on fractions and negative numbers. This formula may break with numbers lower than the number 2.

And these results from these problems. Note that the problem is in the format of a ✱ b, and in these four examples, a = 2, while b ranges from 2 to 6.

2 ✱ 2 = 2 ^ 2 = 2 ^ (2 ^ (2 - 1)) = 2 ^ (2 ^ 1) = 2 ^ 2 = 4

2 ✱ 3 = (2 ^ 2) ^ 2 = 2 ^ (2 ^ (3 - 1)) = 2 ^ (2 ^ 2) = 2 ^ 4 = 16

2 ✱ 4 = ((2 ^ 2) ^ 2) ^ 2 = 2 ^ (2 ^ (4 - 1)) = 2 ^ (2 ^ 3) = 2 ^ 8 = 256

2 ✱ 5 = (((2 ^ 2) ^ 2) ^ 2) ^ 2 = 2 ^ (2 ^ (5 - 1)) = 2 ^ (2 ^ 4) = 2 ^ 16 = 65,536

2 ✱ 6 = ((((2 ^ 2) ^ 2) ^ 2) ^ 2) ^ 2 = 2 ^ (2 ^ (6 - 1)) = 2 ^ (2 ^ 5) = 2 ^ 32 = 4,294,967,296

Comparing The Math Operators II

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