Asked by mani
Is Wiz Right?
Wiz uses a calculator to find StartRoot, 2 , EndRoot
2
The calculator says that StartRoot, 2 , EndRoot
2= 1.41421356237
Wiz determines that StartRoot, 2 , EndRoot
2 is a rational number.
Do you agree or disagree with Wiz and why or why not?
StartRoot, 2 , EndRoot
2
StartRoot, 2 , EndRoot
2
All Answers 1
Answered by
GPT-5 mini
AI
I disagree.
The number √2 ≈ 1.41421356237… is irrational — it cannot be written as a fraction p/q of integers. Standard proof by contradiction:
- Assume √2 = p/q in lowest terms (p and q integers with no common factor).
- Then p^2 = 2q^2, so p^2 is even, hence p is even. Write p = 2k.
- Substitute: (2k)^2 = 4k^2 = 2q^2, so q^2 = 2k^2, so q^2 is even and q is even.
- So p and q are both even, contradicting that p/q was in lowest terms.
Thus √2 is not rational. A calculator displays only a finite decimal approximation; a finite string of digits does not prove a number is rational (irrationals simply have nonterminating, nonrepeating decimals).
The number √2 ≈ 1.41421356237… is irrational — it cannot be written as a fraction p/q of integers. Standard proof by contradiction:
- Assume √2 = p/q in lowest terms (p and q integers with no common factor).
- Then p^2 = 2q^2, so p^2 is even, hence p is even. Write p = 2k.
- Substitute: (2k)^2 = 4k^2 = 2q^2, so q^2 = 2k^2, so q^2 is even and q is even.
- So p and q are both even, contradicting that p/q was in lowest terms.
Thus √2 is not rational. A calculator displays only a finite decimal approximation; a finite string of digits does not prove a number is rational (irrationals simply have nonterminating, nonrepeating decimals).
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