We shall prove this by contradiction. Let P1 = 2 ; P2 = 3 ; P3 = 5 ; . . . . . . . . . . . Pn = m, be the primes in ascending order .suppose there exists a last prime Pn , consider a positive integer “P” = (P1P2P3P4 . . . . . . Pn) + 1 .since P > 1 , some prime number divides “P” . let that number be “A” .since we assumed primes are infinite , “A” must be one among the set {p1 , p2 , p3 , . . . . . . . pn}.So , “A” should divide 1. Here, a contradiction arises. therefore there are infinite number of primes .

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TopNewestHey Sudoku Subbu!The proof is nice but to make it a bit more pleasing to the eye you can use \(\LaTeX\).For example,instead of "Pn" you can get \(P_{n}\).The code is \ ( P_{n} \ ).Just remove the spaces. – Adarsh Kumar · 2 years ago

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– Sudoku Subbu · 2 years ago

i tried that but it remained same as i typed by the by thanks for your appreciationLog in to reply