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Click to Select a Page... 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 Fun With Math Strip Home Page DavesFunStuff Dave's Fun Stuff Find your favorite TV shows with "Let's Watch TV!" Dave's Fun Stuff SDN Media News and More Dave's Fun Stuff Super Birthdays Contact Webmaster

Using Numbers 3 and Higher

We're going to explore the use of the superpowerition ✱ operation with integers "3" and higher. For now, we're restricting our scope to integers 2 and higher for both arguments in the operation a ✱ b until the next page.

In the previous page, you have seen the number two in action. We'll be using this formula a ^ (a ^ (b - 1)) in our examples:

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

The previous examples were when a = 2. Now, let a = 3 and b range from 2 to 6 in the following examples:

3 ✱ 2 = 3 ^ 3 = 3 ^ (3 ^ (2 - 1)) = 3 ^ (3 ^ 1) = 3 ^ 3 = 27

3 ✱ 3 = (3 ^ 3) ^ 3 = 3 ^ (3 ^ (3 - 1)) = 3 ^ (3 ^ 2) = 3 ^ 9 = 19,683

3 ✱ 4 = ((3 ^ 3) ^ 3) ^ 3 = 3 ^ (3 ^ (4 - 1)) = 3 ^ (3 ^ 3) = 3 ^ 27 = 7.6255974 × 1012

3 ✱ 5 = (((3 ^ 3) ^ 3) ^ 3) ^ 3 = 3 ^ (3 ^ (5 - 1)) = 3 ^ (3 ^ 4) = 3 ^ 81 = 4.4342648 × 1038

3 ✱ 6 = ((((3 ^ 3) ^ 3) ^ 3) ^ 3) ^ 3 = 3 ^ (3 ^ (6 - 1)) = 3 ^ (3 ^ 5) = 3 ^ 243 = [CALCULATOR BUSTED]

As you can see, 3 ✱ 6 results in a high number, so high that it busted my calculator! I'll have to use a phone app to get the result!

3 ✱ 6 = ((((3 ^ 3) ^ 3) ^ 3) ^ 3) ^ 3 = 3 ^ (3 ^ (6 - 1)) = 3 ^ (3 ^ 5) = 3 ^ 243 = 8.7189642 × 10¹¹⁵

Now, let's let a = 4 and b range from 2 to 6 in the following examples:

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

4 ✱ 3 = (4 ^ 4) ^ 4 = 4 ^ (4 ^ (3 - 1)) = 4 ^ (4 ^ 2) = 4 ^ 16 = 4,294,967,296

4 ✱ 4 = ((4 ^ 4) ^ 4) ^ 4 = 4 ^ (4 ^ (4 - 1)) = 4 ^ (4 ^ 3) = 4 ^ 64 = 3.4028236 × 1038

4 ✱ 5 = (((4 ^ 4) ^ 4) ^ 4) ^ 4 = 4 ^ (4 ^ (5 - 1)) = 4 ^ (4 ^ 4) = 4 ^ 256 = 1.3407807 × 10154

4 ✱ 6 = ((((4 ^ 4) ^ 4) ^ 4) ^ 4) ^ 4 = 4 ^ (4 ^ (6 - 1)) = 4 ^ (4 ^ 5) = 4 ^ 1024 = 3.2317006 × 10616

As you can see, the numbers are getting exponentially higher, and using at least one number "4" in either of the a or b variables along with another number of "3" or higher results in an answer of at least a billion!

I busted my calculator. Let's see if I can bust my phone app with a = 5 and b ranging from 2 to 6.

5 ✱ 2 = 5 ^ 5 = 5 ^ (5 ^ (2 - 1)) = 5 ^ (5 ^ 1) = 5 ^ 5 = 3,125

5 ✱ 3 = (5 ^ 5) ^ 5 = 5 ^ (5 ^ (3 - 1)) = 5 ^ (5 ^ 2) = 5 ^ 25 = 2.9802322 × 1017

5 ✱ 4 = ((5 ^ 5) ^ 5) ^ 5 = 5 ^ (5 ^ (4 - 1)) = 5 ^ (5 ^ 3) = 5 ^ 125 = 2.3509887 × 1087

5 ✱ 5 = (((5 ^ 5) ^ 5) ^ 5) ^ 5 = 5 ^ (5 ^ (5 - 1)) = 5 ^ (5 ^ 4) = 5 ^ 625 = 7.1821208 × 10436

5 ✱ 6 = ((((5 ^ 5) ^ 5) ^ 5) ^ 5) ^ 5 = 5 ^ (5 ^ (6 - 1)) = 5 ^ (5 ^ 5) = 5 ^ 3,125 = 1.9110125 × 102184

The last operation didn't bust my phone app, but we're getting seriously high numbers now, except for the argument where b = 2.

I'm going to continue on, but let a = 6 to 10 and b remaining at 2. a ^ (a ^ (b - 1)) becomes a ^ (a ^ (2 - 1)) or simply a ^ (a ^ 1) or just a ^ a.

6 ✱ 2 = 6 ^ 6 = 6 ^ ( 6 ^ (2 - 1)) = 6 ^ ( 6 ^ 1) = 6 ^ 6 = 46,656

7 ✱ 2 = 7 ^ 7 = 7 ^ ( 7 ^ (2 - 1)) = 7 ^ ( 7 ^ 1) = 7 ^ 7 = 823,543

8 ✱ 2 = 8 ^ 8 = 8 ^ ( 8 ^ (2 - 1)) = 8 ^ ( 8 ^ 1) = 8 ^ 8 = 16,777,216

9 ✱ 2 = 9 ^ 9 = 9 ^ ( 9 ^ (2 - 1)) = 9 ^ ( 9 ^ 1) = 9 ^ 9 = 387,420,489

10 ✱ 2 = 10 ^ 10 = 10 ^ (10 ^ (2 - 1)) = 10 ^ (10 ^ 1) = 10 ^ 10 = 10,000,000,000

Looks like we cracked the billion mark! That's all I'm going to do with integers 2 and up.

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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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