Why Rational-Radical Domains Need Care
A formula containing both a fraction and a square root has several ways to become undefined. A denominator may be zero, a square root may have a negative radicand, or a square root in a denominator may be zero. These restrictions apply at the same time, so satisfying just one of them is not enough.
The reliable approach is to translate each operation into a condition, combine all the conditions, and then check points where equality or a denominator restriction could change whether a point belongs to the domain. In multivariable formulas, those candidate boundary points may lie on curves or surfaces rather than at isolated numbers.
The Restrictions to Apply
For real-valued formulas, a square root \(\sqrt{A(x,y)}\) is defined when its radicand satisfies \(A(x,y)\geq 0\). A fraction \(A(x,y)/B(x,y)\) is defined only when \(B(x,y)\neq 0\). If a square root is itself in a denominator, then its radicand must be strictly positive: \(\sqrt{A(x,y)}\neq 0\) requires \(A(x,y)>0\).
For example, a formula may contain both \(\sqrt{A(x,y)}\) and a denominator \(B(x,y)\). Its conditions include \(A(x,y)\geq 0\) and \(B(x,y)\neq 0\). If instead \(\sqrt{A(x,y)}\) is in the denominator, the condition is \(A(x,y)>0\), not merely \(A(x,y)\geq 0\).
A rational expression inside a square root requires both types of care. For \(\sqrt{N(x,y)/D(x,y)}\), its conditions are \(D(x,y)\neq 0\) and \(N(x,y)/D(x,y)\geq 0\). The quotient must be defined before its sign can be considered.
A Short Justification for the Rules
These restrictions follow directly from the meanings of the operations. A real square root is the nonnegative number whose square is the radicand, so no real value exists when the radicand is negative. Division by zero is undefined. In particular, if \(\sqrt{A}\) is a denominator, then \(A\geq 0\) is necessary for the square root to be real, and \(A=0\) would make that denominator zero. Together those facts give \(A>0\).
There is also a useful sign rule for a rational radicand. Whenever \(D\neq 0\), the quantities \(N/D\) and \(ND\) have the same sign, because
More directly, \(1/D\) and \(D\) have the same sign when \(D\neq 0\), so multiplying \(N\) by either one gives the same sign. Thus, with \(D\neq 0\), the condition \(N/D\geq 0\) is equivalent to \(ND\geq 0\). The nonzero-denominator condition must remain in the domain description; the product inequality does not replace it.
A Practical Procedure
Locate every square root and every denominator, including denominators nested inside a radicand.
Use a nonnegative radicand for a square root, a nonzero denominator for a fraction, and a strictly positive radicand for a square root in a denominator.
Express the domain as the set of points satisfying all restrictions simultaneously.
Substitute points on equality curves and excluded denominator sets into the original formula, not just a simplified inequality.
Worked Examples
Worked Example: A Square Root with a Separate Denominator
Find the natural domain of \(f(x,y)=\frac{\sqrt{x+y-2}}{x-2y}\).
The square root requires its radicand to be nonnegative:
The denominator must also be nonzero:
Both conditions must hold, so the domain is
Check points on the square-root boundary \(x+y=2\). At \((1,1)\), the radicand is \(1+1-2=0\), and the denominator is \(1-2=-1\), so \(f(1,1)=0/(-1)=0\); the point is included. At \((2,0)\), the radicand is \(2+0-2=0\), and the denominator is \(2-0=2\), so \(f(2,0)=0/2=0\); this point is included too.
Now check \((2,1)\), where the denominator vanishes. Its radicand is \(2+1-2=1\), but its denominator is \(2-2(1)=0\), so the formula is undefined. It is excluded. The boundary equality for the square root is allowed, but the denominator equality is not.
Worked Example: A Rational Radicand
Find the domain of \(g(x,y)=\frac{\sqrt{(x-y)/(x+y)}}{x-1}\).
The inner quotient requires \(x+y\neq 0\). Its value must also be nonnegative to be under a real square root:
When \(x+y\neq 0\), this quotient inequality is equivalent to \((x-y)(x+y)\geq 0\). The outer denominator supplies one more restriction, \(x-1\neq 0\), or \(x\neq 1\). Therefore
Check a point where the numerator of the inner quotient is zero: at \((2,2)\), the quotient is \((2-2)/(2+2)=0/4=0\), and the outer denominator is \(2-1=1\). Thus \(g(2,2)=0/1=0\), so the point is included.
At \((2,0)\), the quotient is \((2-0)/(2+0)=2/2=1\), and the outer denominator is \(2-1=1\). Hence \(g(2,0)=\sqrt{1}/1=1\), so it is included. At \((0,2)\), the quotient is \((0-2)/(0+2)=-2/2=-1\), which has no real square root; the point is excluded. At \((1,0)\), the radicand is \(1\), but the outer denominator is \(1-1=0\), so this point is excluded. Finally, at \((0,0)\), the inner denominator \(x+y\) is zero; the quotient is undefined, even though the product inequality alone gives \((x-y)(x+y)=0\). This confirms why the nonzero condition must be kept.
Worked Example: A Square Root in the Denominator
Find the domain of \(h(x,y)=\frac{\sqrt{(x+y-1)/(x-y+2)}}{\sqrt{9-x^2-y^2}}\).
The rational radicand in the numerator requires \(x-y+2\neq 0\) and \((x+y-1)/(x-y+2)\geq 0\). The denominator is a square root, so its radicand must be strictly positive:
Write \(N=x+y-1\) and \(D=x-y+2\). For \(D\neq 0\), the quotient \(N/D\) is nonnegative when \(N\) and \(D\) have the same sign, or when \(N=0\). This is equivalent to \(ND\geq 0\). The resulting domain is
At \((1,0)\), the numerator of the rational radicand is \(1+0-1=0\), its denominator is \(1-0+2=3\), and the second radicand is \(9-1^2-0^2=8\). Therefore \(h(1,0)=\sqrt{0/3}/\sqrt{8}=0/\sqrt{8}=0\), so the point is included. At \((0,1)\), the first radicand is \((0+1-1)/(0-1+2)=0/1=0\), and the denominator radicand is \(9-0-1=8>0\). Thus \(h(0,1)=0/\sqrt{8}=0\), also included.
At \((3,0)\), the first radicand is \((3+0-1)/(3-0+2)=2/5\), which is positive, but the denominator radicand is \(9-3^2-0^2=0\). The denominator becomes \(\sqrt{0}=0\), so the point is excluded. At \((-1/2,3/2)\), both \(x+y-1\) and \(x-y+2\) equal zero. Their quotient is \(0/0\), which is undefined. The denominator radicand is \(9-1/4-9/4=13/2>0\), but that does not repair the undefined quotient. This point is excluded as well.
Boundary Checks and a Common Mistake
In domain analysis, a boundary equation does not by itself tell you whether its points are included. The original operation determines that. For \(\sqrt{A}\), equality \(A=0\) is allowed. For a denominator \(B\), equality \(B=0\) is forbidden. For a square-root denominator, equality of its radicand to zero is forbidden. When several restrictions meet, check all of them at the candidate point.
A common mistake is to see that a product inequality holds and conclude that the point is in the domain. In the second example, the point \((0,0)\) satisfies \((x-y)(x+y)\geq 0\), since the product is zero. But the original radicand contains \((x-y)/(x+y)=0/0\), which is undefined. The correction is to retain \(x+y\neq 0\) alongside the product inequality. A sign-equivalent inequality does not erase the restriction that made the quotient meaningful.
It is also risky to simplify an expression before recording its restrictions. Algebraic cancellation can hide values where the original formula has a zero denominator. The domain belongs to the original formula: write down its restrictions first, and use algebraic transformations only while preserving those restrictions.
Check Your Understanding
For each question, identify all restrictions before deciding whether a point is in the domain.
- Find the natural domain of \(p(x,y)=\sqrt{2x-y}/(x+y)\).
- For \(q(x,y)=\sqrt{(x+1)/(y-2)}\), what conditions must hold before the formula is defined?
- Does \((1,1)\) belong to the domain of \(r(x,y)=\sqrt{x-y}/(x-1)\)? Explain each restriction.
- Why is the condition \((x+1)(y-2)\geq 0\) alone insufficient to describe the domain of \(q\) in question 2?
- For \(s(x,y)=1/\sqrt{4-x^2-y^2}\), should the radicand condition be nonnegative or strictly positive? Explain.