sqrt(5) is not rational. I will prove this by contradiction. Let’s say that sqrt(5) is rational, that is sqrt(5) = m/n for some m, n in the set of positive integers and m and n are coprime (so that we know it is a unique solution). If we square both sides of the equation, we get 5 = (m/n)^2 = m^2/n^2. Then we can multiply both sides of the equation by n^2 and we get 5n^2 = m^2. The equation is saying that 5 divides m^2 and therefore divides m. This means that some positive integer p exists such that 5p = m. So now 5n^2 = m^2 = (5p)^2 = 25p^2 and then we can divide 5n^2 = 25p^2 by 5 and get n^2 = 5p^2. Then divide both sides by p^2 and obtain 5 = n^2/p^2. Now take the square root of both sides and we get sqrt(5) = n/p. This leads to a contradiction because from the original equation, sqrt(5) = m/n and now we have sqrt(5) = n/p. In the original equation, m is greater than n and in the second equation, n is greater than p. In other words, this is saying p < n < m. n and p are another pair of integers other than m and n that have a quotient of sqrt(5) which contradicts our assumption that m and n are unique.
The square root of 17 is not rational. I will prove this by contradiction. Let’s say that sqrt(17) is rational, that is sqrt(17) = m/n for some m, n in the set of positive integers and m and n are coprime (so that we know it is a unique solution). If we square both sides of the equation, we get 17 = (m/n)^2 = m^2/n^2. Then we can multiply both sides of the equation by n^2 and we get 17n^2 = m^2. The equation is saying that 17 divides m^2 and therefore divides m. This means that some positive integer p exists such that 17p = m. So now 17n^2 = m^2 = (17p)^2 = 289p^2 and then we can divide 17n^2 = 289p^2 by 17 and get n^2 = 17p^2. Then divide both sides by p^2 and obtain 17 = n^2/p^2. Now take the square root of both sides and we get sqrt(17) = n/p. This leads to a contradiction because from the original equation, sqrt(17) = m/n and now we have sqrt(17) = n/p. In the original equation, m is greater than n and in the second equation, n is greater than p. In other words, this is saying p < n < m. n and p are another pair of integers other than m and n that have a quotient of sqrt(17) which contradicts our assumption that m and n are unique.
The square root of 9 is rational and the types of proofs above will not show that the square root of 9 is irrational. This is because if we assume that sqrt(9) = m/n, m, n are in the set of positive integers and are coprime, we can see that 9 = (m/n)^2 = m^2/n^2. If we let m = 3 and n = 1, 3^2/1^2 = 9/1 = 9.
The cubed root of 2 is not a rational number and I will prove this by contradiction. Let’s assume that there exists an m, n such that crt(2) = m/n and m and n are coprime (this means that m and n are the smallest positive integers that exist such that m/n = crt(2). If we cube both sides of the equation, we get 2 = (m/n)^3 = m^3/n^3. We can multiply each side of the equation by n^3 to get 2n^3 = m^3. This means that 2 goes into m^3 and goes into m and therefore there is a positive integer, p such that 2p = m. Now we can substitute 2p in for m in 2n^3 = m^3 and we get 2n^3 = (2p)^3 = 8p^3. Now let’s divide both sides of the equation by 2 and we get n^3 = 4p^3 = 2*2p^3. This means that 2 also divides n^3 and therefore 2 divides n. This goes against the assumption that m and n share no common factors other than 1 since 2 divides both m and n.
Log base 10 of 5 is not rational. I will prove this by contradiction. Let’s assume that log105 is rational, that is log105 = m/n, m and n are positive integers and are coprime. This means that 10^(m/n) = 5. We can raise both sides of the equation to the power of n and we get 5^n = 10^(m/n)^n= 10^m. There are no positive integers, m and n such that 5^n = 10^m. 5 to the n would result in a positive integer that ends with a 5 and 10 to the m would result in a positive integer that ends with a 0.
I cannot see the diagrams on mathematica and I also tried looking up these diagrams and still could not find them.
The first mathematica code took 122.4573370 seconds and the second one took 7.0831204 seconds. There is a big difference probably because since the second one stores the computed results in memory, it takes less time for those values to be used again. Since the first one does not save the computed results in memory, it takes a lot more time for those values to be computed again to be used.
I have inserted both the array and the plot of x1., … , x25 below. The plot shows that the sequence converges to 2. The sequence is infinitely decreasing, but is never less than 2.
