IsQuadraticIrrational

True when $x$ is an irrational root of a quadratic with integer coefficients — exactly the numbers with an eventually periodic continued fraction.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguageQuadraticIrrationalQ✓↗

IsQuadraticIrrational(x)whether $x$ is a quadratic irrational

Typed signatures

Type	Package	Applies
(value) -> boolean	modular	always

Domain: The modular group

Details

• Recognised structurally: a rational affine combination with exactly one irrational Sqrt term, at any rational scale — $\sqrt n$, $3\sqrt2$, $1+\sqrt5$, $(1+\sqrt5)/2$, $1-\sqrt3$
• A rational is not irrational, so it is False, not merely unrecognised
• Wolfram calls this QuadraticIrrationalQ; the name here follows compute-engine's own Is… convention instead
• Every quadratic irrational's continued fraction is eventually periodic (Lagrange's theorem), and conversely — see ContinuedFraction

Examples

Basic #

live ↗

In$\mathrm{IsQuadraticIrrational}(\sqrt{2})$Out$\mathrm{True}$

the golden ratio

live ↗

In$\mathrm{IsQuadraticIrrational}(\frac{1+\sqrt{5}}{2})$Out$\mathrm{True}$

a cube root is algebraic of degree 3

live ↗

In$\mathrm{IsQuadraticIrrational}(\sqrt[3]{2})$Out$\mathrm{False}$

$1+\sqrt2$ is a root of a quadratic with integer coefficients

live ↗

In$\mathrm{IsQuadraticIrrational}(1+\sqrt{2})$Out$\mathrm{True}$

$\sqrt2+\sqrt3$ has degree $4$, not $2$

live ↗

In$\mathrm{IsQuadraticIrrational}(\sqrt{2}+\sqrt{3})$Out$\mathrm{False}$

Scope #

transcendental

live ↗

In$\mathrm{IsQuadraticIrrational}(π)$Out$\mathrm{False}$

rational once simplified

live ↗

In$\mathrm{IsQuadraticIrrational}(\sqrt{4})$Out$\mathrm{False}$

a plain integer

live ↗

In$\mathrm{IsQuadraticIrrational}(5)$Out$\mathrm{False}$

a rational multiple of a surd

live ↗

In$\mathrm{IsQuadraticIrrational}(3\sqrt{2})$Out$\mathrm{True}$

a surd subtracted from a rational

live ↗

In$\mathrm{IsQuadraticIrrational}(1-\sqrt{3})$Out$\mathrm{True}$

Rationals are not irrational at all

live ↗

In$\mathrm{IsQuadraticIrrational}(\frac{3}{2})$Out$\mathrm{False}$

Threads over a list; the integer $2$ is rational

live ↗

In$\mathrm{IsQuadraticIrrational}(\bigl\lbrack\sqrt{2}, \sqrt{3}, 2\bigr\rbrack)$Out$\bigl\lbrack\mathrm{True}, \mathrm{True}, \mathrm{False}\bigr\rbrack$

live ↗

In$\mathrm{IsQuadraticIrrational}(s)$Out$\mathrm{False}$

Possible issues #

a rational is not irrational

live ↗

In$\mathrm{IsQuadraticIrrational}(\frac{3}{4})$Out$\mathrm{False}$

About

Examples	14
Checked against	Wolfram Language 14 agree
Crosswalk	Wolfram Language

See also: ContinuedFraction, Convergents