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You are at the section Fun With Math-Superpowerition: Inventing a Super Power Operation

Integer a > 1 ✱ Decimal b > 1

Up until now, we used the Superpowerition operation for integers.

Now, let's use the operation for decimals, fractions, irrational numbers, constants, pi and more.

On this page, let's make the a in "a ✱ b" an integer > 1 and the b a decimal number > 1.

Let's try out some variables for a using the formula a ^ (a ^ (b - 1)). Let's let b go from 1 to 2 in 0.25 steps.

Let's let a = 2.

2 ✱ 1.00 = 2 ^ (2 ^ (1.00 - 1)) = 2 ^ (2 ^ (0.00)) = 2 ^ 1 = 2
2 ✱ 1.25 = 2 ^ (2 ^ (1.25 - 1)) = 2 ^ (2 ^ (0.25)) ≈ 2 ^ 1.1892071 ≈ 2.2802738
2 ✱ 1.50 = 2 ^ (2 ^ (1.50 - 1)) = 2 ^ (2 ^ (0.50)) ≈ 2 ^ 1.4142135 ≈ 2.6651440
2 ✱ 1.75 = 2 ^ (2 ^ (1.75 - 1)) = 2 ^ (2 ^ (0.75)) ≈ 2 ^ 1.6817928 ≈ 3.2082616
2 ✱ 2.00 = 2 ^ (2 ^ (2.00 - 1)) = 2 ^ (2 ^ (1.00)) = 2 ^ 2 = 4

It seems that when b is a decimal number, the answers are irrational numbers in many cases. So, let's use this formula instead a√(a(ab)) or (a ^ (a ^ b)) ^ (1 / a)

Let's let a = 4.

2 ✱ 1.00 = (2 ^ (2 ^ 1.00)) ^ (1 / 2) = (2 ^ (2.0000000)) ^ (1/2) = (4.0000000) ^ (1/2) = 2
2 ✱ 1.25 = (2 ^ (2 ^ 1.25)) ^ (1 / 2) = (2 ^ (2.3784142)) ^ (1/2) ≈ (5.1996488) ^ (1/2) ≈ 2.2802738
2 ✱ 1.50 = (2 ^ (2 ^ 1.50)) ^ (1 / 2) = (2 ^ (2.8284271)) ^ (1/2) ≈ (7.1029921) ^ (1/2) ≈ 2.6651440
2 ✱ 1.75 = (2 ^ (2 ^ 1.75)) ^ (1 / 2) = (2 ^ (3.3635856)) ^ (1/2) ≈ (10.292957) ^ (1/2) ≈ 3.2082616
2 ✱ 2.00 = (2 ^ (2 ^ 2.00)) ^ (1 / 2) = (2 ^ (4.0000000)) ^ (1/2) = (16.0000000) ^ (1/2) = 4

The answers using the two formulas are the same.

Let's let a = 4 and b = go from 1 to 2 in 0.25 steps, using the formula a ^ (a ^ (b - 1)).

4 ✱ 1.00 = 4 ^ (4 ^ (1.00 - 1)) = 4 ^ (4 ^ (0.00)) = 4 ^ 1 = 4
4 ✱ 1.25 = 4 ^ (4 ^ (1.25 - 1)) = 4 ^ (4 ^ (0.25)) ≈ 4 ^ 1.4142135 ≈ 7.1029926
4 ✱ 1.50 = 4 ^ (4 ^ (1.50 - 1)) = 4 ^ (4 ^ (0.50)) = 4 ^ 2 = 16
4 ✱ 1.75 = 4 ^ (4 ^ (1.75 - 1)) = 4 ^ (4 ^ (0.75)) ≈ 4 ^ 2.8284270 ≈ 50.4525051
4 ✱ 2.00 = 4 ^ (4 ^ (2.00 - 1)) = 4 ^ (4 ^ (1.00)) = 4 ^ 4 = 256

You may have noticed that an exception was found for the expression 4 ✱ 1.50 in which the answer is a rational number of 16.

Let's let a = 2 and b = go from 2 to 3 in 0.25 steps, using the formula a ^ (a ^ (b - 1)).

2 ✱ 2.00 = 2 ^ (2 ^ (2.00 - 1)) = 2 ^ (2 ^ (1.00)) = 2 ^ 2 = 4
2 ✱ 1.25 = 2 ^ (2 ^ (2.25 - 1)) = 2 ^ (2 ^ (1.25)) ≈ 2 ^ 2.3784142 ≈ 5.1996488
2 ✱ 1.50 = 2 ^ (2 ^ (2.50 - 1)) = 2 ^ (2 ^ (1.50)) ≈ 2 ^ 2.8284271 ≈ 7.1029931
2 ✱ 1.75 = 2 ^ (2 ^ (2.75 - 1)) = 2 ^ (2 ^ (1.75)) ≈ 2 ^ 3.3645856 ≈ 10.3000940
2 ✱ 2.00 = 2 ^ (2 ^ (3.00 - 1)) = 2 ^ (2 ^ (2.00)) = 2 ^ 4 = 16

We discover that 4 ✱ 1.00 = 2 ✱ 2.00, and 4 ✱ 1.50 = 2 ✱ 2.00, but due to approximation, 4 ✱ 1.25 is very close to equalling 2 ✱ 1.50, but since I'm using seven decimal places, and due to the nature of irrational numbers, they might indeed be equal.

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Fun With Math-Superpowerition: Inventing a Super Power Operation Main Page Introduction Comparing The Math Operators I Comparing The Math Operators II Using Numbers 3 and Higher 1 ✱ b a ✱ 1 a ✱ 0 0 ✱ b a ✱ -b -a ✱ b -a ✱ -b Formula Summary So Far Integer a > 1 ✱ Decimal b > 1 Integer a > 1 ✱ Decimal 0 < b < 1 Integer a > 1 ✱ Decimal b < 0 Decimal a > 1 ✱ Decimal b > 1 Decimal a > 1 ✱ Decimal 0 < b < 1 Decimal a > 1 ✱ Decimal b < 0 Decimal 0 < a < 1 ✱ Decimal b > 1 Decimal 0 < a < 1 ✱ Decimal 0 < b < 1 Decimal 0 < a < 1 ✱ Decimal b < 0 Decimal -2 < a < 0 ✱ Decimal -1 < b < 2 Formula Summary So Far II Why 2 ✱ 0 is Not 1 Finding The Inverse Operations 1.25 ✱ -10 to 10 1.5 ✱ -10 to 10 1.75 ✱ -10 to 10 2 ✱ -10 to 10 2.5 ✱ -10 to 10 3 ✱ -10 to 10 3.5 ✱ -10 to 10 4 ✱ -10 to 10 4.5 ✱ -10 to 10 5 ✱ -10 to 10 6 ✱ -10 to 10 7 ✱ -10 to 10 8 ✱ -10 to 10 9 ✱ -10 to 10 10 ✱ -10 to 10
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