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 P: 258 Let all positive whole number be written as a sum of power of two: $$1 = 2^0$$ $$2 = 2^1$$ $$3 = 2^0 + 2^1$$ $$4 = 2^2$$ $$5 = 2^0 + 2^2$$ $$6 = 2^1 + 2^2$$ $$7 = 2^0 + 2^1 + 2^2$$ $$8 = 2^3$$ $$9 = 2^0 + 2^3$$ $$10 = 2^1 + 2^3$$ $$11 = 2^0 + 2^1 + 2^3$$ $$12 = 2^2 + 2^3$$ $$13 = 2^0 + 2^2 + 2^3$$ $$14 = 2^1 + 2^2 + 2^3$$ $$15 = 2^0 + 2^1 + 2^2 + 2^3$$ $$16 = 2^4$$ $$17 = 2^0 + 2^4$$ $$18 = 2^1 + 2^4$$ $$19 = 2^0 + 2^1 + 2^4$$ $$20 = 2^2 + 2^4 etc...$$ We know by Tchebychef: for m > 1 there is at least one prime p such that m < p < 2m if we put m = 2^n, n a natural > 0 ==> 2^n < p < 2^(n+1) We know by Gauss: the number of primes from 1 to x is approximately =$$\frac{x}{ln(x)}$$ The number of primes between $$2^n$$ and $$2^{n+1}$$ is approximately = $$\frac{2^{n+1}}{ln(2^{n+1})} - \frac{2^n}{ln(2^n)}$$ = $$\frac{2^n}{ln(2)}*(\frac{2}{n+1} - \frac{1}{n})$$ = $$\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})$$ and we know that $$2^n > n^2$$ for $$\geq 3$$ ==> $$\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})$$ diverges for n --> $$\infty$$ for n increasing the Tchebychef's theorem underestimates the amount of primes