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Exercise 2195Number theory › PrimesM1Added 3 Nov 2026

Chebyshev's bounds

#Chebyshev#bounds#prime counting

Statement

State Chebyshev's bounds on π(x)\pi(x) and apply them at x=1000x=1000.

Hint

Chebyshev: a x/ln⁡x≤π(x)≤b x/ln⁡xa\,x/\ln x\le\pi(x)\le b\,x/\ln x for constants a<ba<b (e.g. a=0.9a=0.9, b=1.1b=1.1).

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Solution

Bounds. There exist constants 0<a<1<b0<a<1<b such that for xx large enough:
axln⁡x≤π(x)≤bxln⁡x.a\frac{x}{\ln x}\le\pi(x)\le b\frac{x}{\ln x}.

Apply at x=1000x=1000. 1000/ln⁡1000≈1451000/\ln1000\approx145. So π(1000)\pi(1000) is within a constant factor of 145145; actual value is 168168.

Conclusion. Chebyshev's bounds prove that π(x)\pi(x) grows like x/ln⁡xx/\ln x up to a constant, a first step toward the prime number theorem.