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  • Jesse Sakari Hyttinen

Numbers game - Jesse Hyttinen

Updated: Apr 21, 2021


Hi and welcome to the Matladpi blog!


How would you prove that


"The number one can be represented as a sequence of infinitely many irrational numbers, where every number is different and the numbers are multiplied by each other" ?


One answer:


1 = 2 - 1 = (2^(1/2))^2 - 1^2

= (2^(1/2) - 1)(2^(1/2) + 1)

= (2^(1/4) - 1)(2^(1/4) + 1)(2^(1/2) + 1)

= (2^(1/8) - 1)(2^(1/8) + 1)(2^(1/4) + 1)(2^(1/2) + 1)

= (2^(1/2) + 1)(2^(1/4) + 1)(2^(1/8) + 1)...

...(2^(2^(-n)) + 1)(2^(2^(-n)) - 1)

, n is a whole number larger than zero, and can be arbitrarily large.


Speaking of multiplication, how would you consider the following:


M! <= M^M


for every whole number M greater than zero.


M! is equal to M^M when M = 1, but M! with values M > 1 will never reach the number M^M.


You may see that this is true, but how would you prove it?


The answer:


For M = 1, M! = 1! = 1 = 1^1 = M^M.


For M > 1:


M! = M(M - 1)(M - 2)(M - 3)...2 * 1 < M^M

ln(M!) = lnM + ln(M - 1) + ln(M - 2) +...+ln2 + ln1 < MlnM, the direction of the relation sign stays the same, as the function f(x) = ln(x) increases continuously in the region x > 1.

ln(M!) - MlnM < 0

(lnM - lnM) + (ln(M - 1) - ln(M)) +...+ (ln2 - lnM) + (ln1 - lnM) < 0

From this you already can see that the relation is true due to the nature of lnx, but:


lna - lnb = ln(a/b)


=> ln1 + ln((M - 1)/M) + ln((M - 2)/M)+...+ ln(2/M) +ln(1/M) < 0


Every term other than ln1 = 0 on the left side is smaller than zero, as


0 < m < M

(M - m)/M < 1

a < 1

lna < 0, lnx also increases continuously in the region 0 < x < 1


There are now M - 1 terms with a negative value, so the whole sum is a negative value and thus the relation is right. In other words:


M! <= M^M.


I am Jesse Sakari Hyttinen and I will see you in the next post!




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