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Adapt the proof in Example 4 in Section 1.7 to prove that if\(n = abc\), where \(a, b, and c\)are positive integers, then\(a \le \sqrt[3]{n}, b \le \sqrt[3]{n}, or c \le \sqrt[3]{n}\).

Short Answer

Expert verified

\(a \le \sqrt[3]{n},b \le \sqrt[3]{n}\)or \(c \le \sqrt[3]{n},\)is true.

Step by step solution

01

Introduction

Positive integral values are those which do not have a negative sign with them and carry a positive sign.

02

Proof using contradiction

Let\(n = abc\), where a, b and c are positive integral values.

Prove that\(a \le \sqrt[3]{n},b \le \sqrt[3]{n}\)or \(c \le \sqrt[3]{n}\)

If it is not true that

\(a \le \sqrt[3]{n},b \le \sqrt[3]{n}\)or \(c \le \sqrt[3]{n},\) then

\(a > \sqrt[3]{n},b > \sqrt[3]{n}\)or \(c > \sqrt[3]{n}\).

Multiplying these inequalities of positive numbers together, it is obtained

\(\begin{aligned}{}abc &< \sqrt[3]{n} \cdot \sqrt[3]{n} \cdot \sqrt[3]{n}\\ < {\left( {\sqrt[3]{n}} \right)^3}\\ &< n\end{aligned}\)

This gives contradiction to the assumption\(n = abc\).

Therefore\(a \le \sqrt[3]{n},b \le \sqrt[3]{n}\)or \(c \le \sqrt[3]{n}\)is true.

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