Missing Rule of Powers
- 1. Powers and Exponents : The Missing Law Allan Cacdac acacdac@gmail.com May 2, 2018 Introduction ”In the beginning was the Word, and the Word was with God, and the Word was God...”. We all have different beliefs regarding this but I, for one, have faith in God. When the word One or the number 1 is used, I would always relate this to our Creator. The creation is I know a different topic but I have to mention it as this was how I came to explore what I have discovered. Hoping to provide this insight to everyone, if not all, at least to One person. So here’s a phrase to remember ... One word, One God and ... One Number. And thus begin our journey to the beginning of numbers that led us to the discovery of the Missing Law of exponents. One = 1 1 The Laws of Exponents First , let’s cover the Laws of Exponents Law of Exponents Application x0 = 1 180 = 1 x−m = 1 xm 18−2 = 1 182 xm+n = xm .xn 182+3 = 182 .183 = 185 (xm )n = xm+n (182 )3 = 182.3 = 186 (xy)m = xm .ym (18.5)2 = 182 .52 = 8100 x y m = xm ym 18 5 2 = 18 5 2 = 182 5 2 xm xn = xm−n 182 183 = 182−3 = 18−1 = 1 18 Please go ahead and grab your favorite math book and search the same references as the table above. Something was missing from my perspective since my first time learning about Powers and Exponents. After so many years, I finally found the answer. Below is what I think and believe should be included. Let’s call the Missing Law of Exponents for the sake of presentation. 1
- 2. Missing Law of Exponent Application xm−1 1 + xm−1 2 + ... + xm−1 x = xm 25 = 25−1 1 + 25−1 2 25 = 24 + 24 32 = 32 38 = 38−1 1 + 38−1 2 + 38−1 3 38 = 37 + 37 + 37 6561 = 6561 74 = 74−1 1 + 74−1 2 + 74−1 3 + 74−1 4 + 74−1 5 + 74−1 6 + 74−1 7 74 = 73 + 73 + 73 + 73 + 73 + 73 + 73 2401 = 2401 5−3 = 5−3−1 1 + 5−3−1 2 + 5−3−1 3 + 5−3−1 4 + 5−3−1 5 = 5−4 + 5−4 + 5−4 + 5−4 + 5−4 1 53 = 1 53 2 In the Beginning We will start of to define, x i=1 1i = x This states that x is exactly the number of terms on the right hand side. We do not need to elaborate how we can simplify in different ways because this is primarily what we need for now and refer to go further. number of terms = x 0 = zero term so x will not have a value of 0 11 = 1 11 + 12 = 2 11 + 12 + 13 = 3 11 + 12 + 13 + 14 = 4 and so on so it is suffice to write 11 + 12 + ... + 1x = x It is pretty simple but looks elegant, right? But indeed, this is something that shouldn’t be just pushed aside. From this, we will move on to complex but still simple equations. 3 Go forth and multiply 3.1 Squares Our next step is to multiply the value of x by itself to both sides of the equation. x i=1 1i = x Multiply x on both sides. 2
- 3. x. x i=1 1i = x.x So we will derive to x i=1 xi = x2 Examples : 1 + 1 = 2 2 . (1 + 1) = 2 . 2 22 = 2 + 2 1 + 1 + 1 = 3 3 . (1 + 1 + 1) = 3 . 3 3 + 3 + 3 = 32 1 + 1 + 1 + 1 = 4 4 . (1 + 1 + 1 + 1) = 4 . 4 4 + 4 + 4 + 4 = 42 3.2 Cubes and beyond We will repeat same steps as previous to multiply the value of x by itself to both sides of the equation. x . x i=1 xi = x2 . x So we will deriveto x i=1 x2 i = x3 Examples : 22 + 22 = 23 32 + 32 + 32 = 33 42 + 42 + 42 + 42 = 43 Repeat same step. Multiply x on both sides. x . x i=1 x2 i = x3 .x So we will derive as x i=1 x3 i = x4 Examples : 3
- 4. 23 + 23 = 24 33 + 33 + 33 = 34 43 + 43 + 43 + 43 = 44 If we keep doing the same, you’ll notice there’s a pattern and we will derive to a new summation that will apply to all xm . That’s how we derived to this Missing Law of Exponents. xm−1 1 + xm−1 2 + ... + xm−1 x = xm Furthermore, we can now use Summation that is equivalent to the Missing Law of Exponents presented above. x i=1 xm−1 i = xm 4 Power Rules Some of us have wondered how did we conclude that any number raised to 0 is 1. But as we have initially stated above, if x = 0 ( no terms), then there was no term originally to work with. How about for values other than 0 ? Here is how we apply the Missing Law to show that any number , x0 = 1. 50 = 50−1 1 + 50−1 2 + 50−1 3 + 50−1 4 + 50−1 5 = 5−1 + 5−1 + 5−1 + 5−1 + 5−1 = 1 5 + 1 5 + 1 5 + 1 5 + 1 5 = 5 5 50 = 1 70 = 70−1 1 + 70−1 2 + 70−1 3 + 70−1 4 + 70−1 5 + +70−1 6 + +70−1 7 = 7−1 + 7−1 + 7−1 + 7−1 + 7−1 + 7−1 + 7−1 = 1 7 + 1 7 + 1 7 + 1 7 + 1 7 + + 1 7 + + 1 7 = 7 7 70 = 1 4
- 5. 4.1 Beyond computing capability Expanding numbers and equations beyond the limitation of computing. Examples : 2100 = 299 + 299 3345 = 3344 + 3344 + 3344 712345 = 712344 + 712344 + 712344 + 712344 + 712344 + 712344 + 712344 .... A calculator will probably overflow if you try to verify some of the numbers or equations as a whole but we understand the pattern and can recognize the validity of the equations. You can probably play around with the numbers as high as you can and no computer in the world can verify your equation but we do know it is correct. 12345678909876543210 = 12345678909876543209 + 12345678909876543209 + ..... + 12345678909876543209 Imagine writing by hand with this kind of equation, we need to somehow present in such a way that we do not need to write the entire terms. But how? Let’s move on. 5 Simplifying Terms So now you are probably wondering what we can do for huge number of terms. We will provide the examples first this time to better grasp the equation that we will be deriving to. Example : 79 = 78 + 78 + 78 + 78 + 78 + 78 + 78 We can simplify this in different ways. 79 = 1 . 78 + (7 − 1)(78 ) 79 = 2 . 78 + (7 − 2)(78 ) 79 = 3 . 78 + (7 − 3)(78 ) 79 = 4 . 78 + (7 − 4)(78 ) 79 = 5 . 78 + (7 − 5)(78 ) 79 = 6 . 78 + (7 − 6)(78 ) Did you notice the pattern? Great. We will now assign variables to those numbers so it will apply to all xy . xy = z . xy−1 + (x − z)(xy−1 ) You can play around some more with this formula and do further study. 5
- 6. 6 Equations, equations and more From a simple summation of 1s, we are able to further expand and explore new equations. It may be a long process how we can derive new ones but they are worth it. In addition, I was able to derive another formula below. But perhaps it can be a great practice for those who are interested. We will cut to the chase before all of you gets bored. Here is the last one to wrap things up... From xy = x i=1 xy−1 i To xy = x + (x − 1) y−1 i=1 xi or simply xy = x + (x − 1)(x1 + x2 + ... + xy−1 ) Examples : 25 = 21 + (2 − 1)(21 + 22 + 23 + 24 ) 29 = 21 + 21 + 22 + 23 + 24 39 = 31 + (3 − 1) (31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 ) 39 = 31 + (2) (31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 ) 7 More discoveries and exploration If you have come this far, I really appreciate for reading this far. So please feel free to reach out to me via email if interested. I have more I’d like to share that I think has not published yet and believe more to be discovered. 6