Proof of Bertrand's postulate: Difference between revisions - Wikipedia

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where the product is taken over all ''prime'' numbers <math>p</math> less than or equal to the real number <math>x</math>.

For all real numbers <math>x\ge 3</math>, <math>x\#<2^{2x-3}. < 4^x</math>.

'''Proof:'''

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:<math>n\# = (2m-1)\# < m\#\cdot2^{2m-2} < 2^{2m-3}2^{2m-2} = 2^{4m-5} = 2^{2n-3}.</math>

* If <math> n=2m </math> is even and <math>n\ge6,</math> then by the above estimate and the induction assumption, since <math>m\ge3</math> and <math>n-1<n</math> it is

:<math>n\# = (n-1)\# < 2^{2(n-1)-3} < 2^{2n-3}.</math>.

Only <math>x#<4^x</math> is used in the proof.

==Proof of Bertrand's Postulate==