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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