Sieve of Eratosthenes

A sieve is a way of weeding out composites with simple arithmetic, for example additions. We shall use the sieve of Eratosthenes to find all primes up to 50. Firstly we write all the numbers from $1$ to $50$. We cross out 1 as that is not a prime. The number $2$ is the first prime and we cross out every second number after it thusly: $[\require{cancel}\xcancel{1},$ $2,$ $3,$ $\xcancel{4},$ $5,$ $\xcancel{6},$ $7,$ $\xcancel{8},$ $9,$ $\xcancel{10},$ $11,$ $\xcancel{12},$ $13,$ $\xcancel{14},$ $15,$ $\xcancel{16},$ $17,$ $\xcancel{18},$ $19,$ $\xcancel{20},$ $21.$ $\xcancel{22},$ $23,$ $\xcancel{24},$ $25,$ $\xcancel{26},$ $27,$ $\xcancel{28},$ $29,$ $\xcancel{30},$ $31,$ $\xcancel{32},$ $33,$ $\xcancel{34},$ $35,$ $\xcancel{36},$ $37,$ $\xcancel{38},$ $39,$ $\xcancel{40},$ $41,$ $\xcancel{42},$ $43,$ $\xcancel{44},$ $45,$ $\xcancel{46},$ $47,$ $\xcancel{48},$ $49,$ $\xcancel{50}]$.

Our next prime is the first to the right of $2$. That is $3$. We then cross out every third number after $3$: $[\require{cancel}\xcancel{1},$ $2,$ $3,$ $\xcancel{4},$ $5,$ $\xcancel{6},$ $7,$ $\xcancel{8},$ $\xcancel{9},$ $\xcancel{10},$ $11,$ $\xcancel{12},$ $13,$ $\xcancel{14},$ $\xcancel{15},$ $\xcancel{16},$ $17,$ $\xcancel{18},$ $19,$ $\xcancel{20},$ $\xcancel{21}.$ $\xcancel{22},$ $23,$ $\xcancel{24},$ $25,$ $\xcancel{26},$ $\xcancel{27},$ $\xcancel{28},$ $29,$ $\xcancel{30},$ $31,$ $\xcancel{32},$ $\xcancel{33},$ $\xcancel{34},$ $35,$ $\xcancel{36},$ $37,$ $\xcancel{38},$ $\xcancel{39},$ $\xcancel{40},$ $41,$ $\xcancel{42},$ $43,$ $\xcancel{44},$ $\xcancel{45},$ $\xcancel{46},$ $47,$ $\xcancel{48},$ $49,$ $\xcancel{50}]$.

The next prime is read off as $5$ and we cross out every fifth number: $[\require{cancel}\xcancel{1},$ $2,$ $3,$ $\xcancel{4},$ $5,$ $\xcancel{6},$ $7,$ $\xcancel{8},$ $\xcancel{9},$ $\xcancel{10},$ $11,$ $\xcancel{12},$ $13,$ $\xcancel{14},$ $\xcancel{15},$ $\xcancel{16},$ $17,$ $\xcancel{18},$ $19,$ $\xcancel{20},$ $\xcancel{21}.$ $\xcancel{22},$ $23,$ $\xcancel{24},$ $\xcancel{25},$ $\xcancel{26},$ $\xcancel{27},$ $\xcancel{28},$ $29,$ $\xcancel{30},$ $31,$ $\xcancel{32},$ $\xcancel{33},$ $\xcancel{34},$ $\xcancel{35},$ $\xcancel{36},$ $37,$ $\xcancel{38},$ $\xcancel{39},$ $\xcancel{40},$ $41,$ $\xcancel{42},$ $43,$ $\xcancel{44},$ $\xcancel{45},$ $\xcancel{46},$ $47,$ $\xcancel{48},$ $49,$ $\xcancel{50}]$.

We read off $7$ as the next prime and cross out every seventh number: $[\require{cancel}\xcancel{1},$ $2,$ $3,$ $\xcancel{4},$ $5,$ $\xcancel{6},$ $7,$ $\xcancel{8},$ $\xcancel{9},$ $\xcancel{10},$ $11,$ $\xcancel{12},$ $13,$ $\xcancel{14},$ $\xcancel{15},$ $\xcancel{16},$ $17,$ $\xcancel{18},$ $19,$ $\xcancel{20},$ $\xcancel{21}.$ $\xcancel{22},$ $23,$ $\xcancel{24},$ $\xcancel{25},$ $\xcancel{26},$ $\xcancel{27},$ $\xcancel{28},$ $29,$ $\xcancel{30},$ $31,$ $\xcancel{32},$ $\xcancel{33},$ $\xcancel{34},$ $\xcancel{35},$ $\xcancel{36},$ $37,$ $\xcancel{38},$ $\xcancel{39},$ $\xcancel{40},$ $41,$ $\xcancel{42},$ $43,$ $\xcancel{44},$ $\xcancel{45},$ $\xcancel{46},$ $47,$ $\xcancel{48},$ $\xcancel{49},$ $\xcancel{50}]$.

Since the next prime $11$ is greater than the square root of $50$ we are done and what remains are the primes less than $50$.