(4) Let n ∈ℕ. Show that if the smallest prime factor p of n satisfies p > n^1/3, then n/p is eit

Name: (4) Let n ∈ℕ. Show that if the smallest prime factor p of n satisfies p > n^1/3, then n/p is either prime or 1.
Uploaded: 2023-10-15T21:29:02-08:00
Duration: 4 min 2 s
Channel: Mengchun Cai
Description: (4) Let n ∈ℕ. Show that if the smallest prime factor p of n satisfies p > n^1/3, then n/p is either prime or 1.

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(4) Let $n \in \mathbb{N}$. Show that if the smallest prime factor $p$ of $n$ satisfies $p > n^{1/3}$, then $n/p$ is either prime or 1.

          (4) Let $n \in \mathbb{N}$. Show that if the smallest prime factor $p$ of $n$ satisfies $p > n^{1/3}$, then $n/p$ is either prime or 1.

(4) Let n ∈ℕ. Show that if the smallest prime factor p of n satisfies p > n^1/3, then n/p is either prime or 1.

Added by Christopher C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Mengchun Cai

10/15/2023

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Step 1: Assume that n/p is not prime. Show more…

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(4) Let n ∈ N. Show that if the smallest prime factor p of n satisfies p > n1/3, then n/p is either prime or 1.

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(4) Let $n \in \mathbb{N}$. Show that if the smallest prime factor $p$ of $n$ satisfies $p > n^{1/3}$, then $n/p$ is either prime or 1.

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