Primes are infinite

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 .

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Hey 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 - 6 years, 5 months ago

i tried that but it remained same as i typed by the by thanks for your appreciation

sudoku subbu - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$