Rational numbers can be written in the following form.Product[p[n]^z[n], {n, 1, Infinity}]Where p[n] is the nth prime number and z[n] is an integer.
>>16542433WOAH! Thanks for your insightful thread about prime number decomposition that every child in elementary school knows about.
The seventh root of nine is irrational.Proof:9^(1/7) = 3^(2/7)p[1] = 2p[2] = 3p[3] = 5p[4] = 7p[5] = 11p[6] = 13p[7] = 17et ceteraz[1] = 0z[2] = 2/7 <> integerz[3] = 0z[4] = 0z[5] = 0z[6] = 0z[7] = 0et cetera
>>16542453The second root of four is irrational.Proof:4^(1/2) = 3^(log(2)/log(3))p[1] = 2p[2] = 3p[3] = 5p[4] = 7p[5] = 11p[6] = 13p[7] = 17et ceteraz[1] = 0z[2] = log(2)/log(3) <> integerz[3] = 0z[4] = 0z[5] = 0z[6] = 0z[7] = 0et cetera
>>16542433Don't forget there's some N such that z[n]=0 for all n>N.
>>16542472you cheatedthe proof only works, if the number is in simplest termsand 3^(log(2)/log(3)) isn't in simplest termssince it's equal to 2
>>16542433>negative numbers are irrational
>>16542433ITT: OP conflates "can" with "must".
>>16542472The ninth root of seven is irrational.Proof:p[1] = 2p[2] = 3p[3] = 5p[4] = 7p[5] = 11p[6] = 13p[7] = 17et ceteraz[1] = 0z[2] = 0z[3] = 0z[4] = 1/9 is rational but not an integerz[5] = 0z[6] = 0z[7] = 0et cetera
>>16542708positive rational numbers can be written in that form
>>16543072Not what OP said
>>16543079>Not what OP saidthe sign of a rational number is trivial
>>16543068>>16542453In other words, both 7^(1/9) and 9^(1/7) are irrational.
