# A New Proof of Euclid's Theorem

Here is another proof of the infinitude of primes.

a_{N} = \sum_{i=1} \frac{1}{p_{i}} - \sum_{{i}<{j}} \frac{1}{p_{i}p_{j}} + \sum_{{i}<{j}<{k}} \frac{1}{p_{i}p_{j}p_{k}} - \ldots + (-1)^{N+{1}}\frac{1}{p_{1}p_{2}· \ldots ·p_{N}}

a_{N} is simply

a_{N} = 1 - \prod_{i={1}}^{N}(1 - \frac{1}{p_{i}})

which shows that 0 \lt a_{N} \lt 1, since each factor is positive and is less than 1.

Assume that the number of primes is finite and list them all p_{1}, p_{2}, \ldots, p_{N}. Let x \gt 1 be a real number. The number of integers in the interval \left[ 1, x\right]  can be expressed via the Inclusion-Exclusion principle. Let for i = 1, \ldots, N, A_{i} denote the set of positive integers in \left[ 1, x\right] that are divisible by p_{i}. The cardinality of \cup A_{i} (which is of course the set of all integers in the interval) is given by

\lfloor x \rfloor = {1} + \sum_{i=1} \bigg\lfloor \frac{x}{p_{i}} \bigg\rfloor - \sum_{{i}<{j}} \bigg\lfloor \frac{x}{p_{i}p_{j}}\bigg\rfloor + \sum_{{i}<{j}<{k}} \bigg\lfloor \frac{x}{p_{i}p_{j}p_{k}}\bigg\rfloor - \ldots + (-1)^{N+{1}} \bigg\lfloor\frac{x}{p_{1}p_{2}· \ldots ·p_{N}} \bigg\rfloor.

However,

\lim_{x\rightarrow \infty}x^{-{1}}\bigg\lfloor\frac{x}{t}\bigg\rfloor = \frac{1}{t}.

Which implies a contradiction:

1 \gt a_{N} = \sum_{i=1} \frac{1}{p_{i}} - \sum_{{i}<{j}} \frac{1}{p_{i}p_{j}} + \sum_{{i}<{j}<{k}} \frac{1}{p_{i}p_{j}p_{k}} - \ldots + (-1)^{N+{1}}\frac{1}{p_{1}p_{2}· \ldots ·p_{N}} = 1.

### References

• J. P. Pinasco, New Proofs of Euclid's and Euler's Theorems, The American Mathematical Monthly, v 116, n 2 (2009), p 172-173