Which statement is correct about \(\sqrt[3]{4}\)?
Answer and explanation
Correct answer: It is irrational
Assume \(\sqrt[3]{4}\) is rational: \(\sqrt[3]{4}=p/q\) in lowest terms (gcd\((p,q)=1\)). Cubing both sides gives \(4q^{3}=p^{3}\). Hence \(p^{3}\) is even so \(p\) is even; write \(p=2k\). Substituting yields \(4q^{3}=8k^{3}\) ⇒ \(q^{3}=2k^{3}\), so \(q\) is even too, contradicting gcd\((p,q)=1\). Therefore \(\sqrt[3]{4}\) is irrational. Option B (2) is incorrect because \(2^{3}=8\), not 4; C and D are also obviously wrong. Exam tip: for cube roots of integers, check whether the radicand is a perfect cube — if not, the cube root is irrational.
Frequently asked questions
What is the correct answer to this question?
It is irrational
Why is this the correct answer?
Assume \(\sqrt[3]{4}\) is rational: \(\sqrt[3]{4}=p/q\) in lowest terms (gcd\((p,q)=1\)). Cubing both sides gives \(4q^{3}=p^{3}\). Hence \(p^{3}\) is even so \(p\) is even; write \(p=2k\). Substituting yields \(4q^{3}=8k^{3}\) ⇒ \(q^{3}=2k^{3}\), so \(q\) is even too, contradicting gcd\((p,q)=1\). Therefore \(\sqrt[3]{4}\) is irrational. Option B (2) is incorrect because \(2^{3}=8\), not 4; C and D are also obviously wrong. Exam tip: for cube roots of integers, check whether the radicand is a perfect cube — if not, the cube root is irrational.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.
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