Can you prove #log_10 2 # is irrational?

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Rhys Share
Dec 2, 2017

Answer:

We can prove this by contradiction;

Explanation:

if #x=QQ# then #x=p/q # for #p,q in ZZ#

so hence let, #log_10 2 =p/q #
then #2 = 10^(p/q)#

so hence # 2^q = 10^p #

#=> 2^q = 5^p 2^p #

#=> 2^(q-p) = 5^p#

we know #log_10 1 < log_10 2 < log_10 10#

so hence # 0 < log_10 2 < 1#

so hence #0 < p/q < 1#

# p < q #

so #q-p > 0 # and let #m=q-p#

But we can see that #2^m # is even and #5^p# is odd, hence there are no values for # p and q# that satisify this

#=># Hence proven by contradiction

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