Algebraic Laws and Valid Transformation

Justify algebraic transformations, preserve domains and solution sets, and detect invalid steps in symbolic arguments.

Why this matters

Algebra is not the art of moving symbols across an equals sign. It is the study of operations that preserve a stated mathematical claim. A line can look plausible, use familiar notation, and still change the domain or the solution set. Dividing by an expression that might be zero can discard solutions; squaring can introduce them; taking a square root can silently select one sign; and canceling a factor can erase the point at which the original expression was undefined.

Those failures become more dangerous as problems become harder. In a routine linear equation, an invalid step may merely give a wrong number. In a parameter problem, the same step may erase an entire family of cases. In proof, it can turn a one-way implication into a false equivalence. In numerical or symbolic software, it can produce a simplified formula whose generic meaning is correct while its exceptional inputs have been lost.

This chapter develops a disciplined replacement for “do the same thing to both sides.” At every line, ask four questions:

  1. Universe: In which number system are the variables allowed to live?
  2. Domain: For which inputs are all expressions in this line defined?
  3. Direction: Is the new line equivalent to the preceding line, or merely a necessary consequence?
  4. Exception audit: Could the operation divide by zero, merge signs, or use a law outside its hypotheses?

The goal is not slower algebra. It is reliable algebra: a small set of laws, applied quickly because their conditions are understood.

What you should be able to do

After completing this chapter, you should be able to:

  • state the field laws and order laws used in elementary real algebra;
  • annotate a transformation with the law or operation that licenses it;
  • derive useful rules such as \(0a=0\), \((-1)a=-a\), and cancellation rather than treating them as unexplained tricks;
  • distinguish equality of values, equality of functions, identities, equations, implications, and equivalences;
  • preserve excluded values when simplifying rational expressions;
  • decide when addition, multiplication, division, squaring, square-rooting, and substitution preserve an equation’s solution set;
  • use the zero-product property and explain why it depends on the absence of zero divisors in the real and complex numbers;
  • construct decisive counterexamples to false laws involving powers, roots, fractions, and inequalities;
  • classify parameter-dependent equations without canceling away exceptional cases;
  • audit a proposed solution by identifying its earliest invalid step; and
  • use SymPy to test identities and candidates while retaining responsibility for assumptions, domains, and logical direction.

Prerequisites

You should be comfortable with set notation, implications, equivalences, and counterexamples from Sets, intervals, statements, and quantifiers. You should also know the nesting and closure properties of the real and complex number systems, the meaning of principal square root, and why \(\sqrt{x^2}=|x|\) from The real number system and its extensions.

Five-minute retrieval warm-up

  1. State the domain over \(\mathbb R\) of \(\dfrac{x+1}{x^2-4}\).
  2. Is \(x=3\Rightarrow x^2=9\) reversible over \(\mathbb R\)? Give a reason.
  3. Simplify \(\sqrt{(x-2)^2}\) for real \(x\).
  4. Negate: “Every value of \(a\) makes the equation solvable.”
  5. Give one counterexample to the claim that the quotient of two integers is an integer.
  1. \(\mathbb R\setminus\{-2,2\}\).
  2. No. \(x=-3\) satisfies \(x^2=9\) but not \(x=3\).
  3. \(|x-2|\).
  4. “There exists a value of \(a\) for which the equation is not solvable.”
  5. For example, \(1/2\notin\mathbb Z\).

1. Equality, substitution, and replacement

An equality \(A=B\) asserts that two expressions denote the same object under the stated variable assignments. Equality has four basic logical properties:

  • reflexive: \(A=A\);
  • symmetric: if \(A=B\), then \(B=A\);
  • transitive: if \(A=B\) and \(B=C\), then \(A=C\);
  • substitutive: if \(A=B\), either may replace the other inside any context in which both resulting expressions are defined.

Substitution is the engine of algebra. From \(A=B\) we obtain \(A+C=B+C\) and \(AC=BC\) because equal inputs may be substituted into the functions \(t\mapsto t+C\) and \(t\mapsto tC\). We obtain \(A^2=B^2\) by substitution into \(t\mapsto t^2\). The reverse statement, however, is a separate question: \(A^2=B^2\) implies \(A=B\) or \(A=-B\), not necessarily \(A=B\).

The phrase “do the same thing to both sides” is safe only after asking whether the chosen function is reversible on the values under consideration. Addition of a fixed number has inverse subtraction and is reversible. Multiplication by a known nonzero number has inverse division and is reversible. Squaring on all of \(\mathbb R\) is not one-to-one and is therefore not reversible without sign information.

Example 1: annotate every line of a simplification

Simplify \(3(2x-5)-2(x+4)\) and justify the transformation.

[ \[\begin{aligned} 3(2x-5)-2(x+4) &=6x-15-2x-8 &&\text{distributive law twice}\\ &=(6x-2x)+(-15-8) &&\text{associative and commutative laws}\\ &=4x-23 &&\text{integer arithmetic and }ax-bx=(a-b)x. \end{aligned}\]

]

Every line has the same value for every real or complex \(x\). No denominator, radical, or variable exponent changes the domain, so the equality is an unconditional identity. A quick check at \(x=2\) gives \(-15\) in both the original and final forms. One numerical check does not prove the identity, but it can catch arithmetic errors.

Replacement checkpoint

From \(u=v\), which conclusion is not always reversible over the reals?

A. \(u+7=v+7\) B. \(5u=5v\) C. \(u^2=v^2\) D. \(u-3=v-3\)

C. Equality implies equality of squares, but the reverse loses sign: \(2^2=(-2)^2\) although \(2\ne-2\). The other operations have inverse operations because the multiplier \(5\) is nonzero.

2. Field laws and rules derived from them

The real and complex numbers are fields. For \(a,b,c\) in a field, the following laws hold.

Structure Laws
Addition \(a+b=b+a\), \((a+b)+c=a+(b+c)\), \(a+0=a\), \(a+(-a)=0\)
Multiplication \(ab=ba\), \((ab)c=a(bc)\), \(a\cdot1=a\), and if \(a\ne0\), \(aa^{-1}=1\)
Connection \(a(b+c)=ab+ac\)

A field also satisfies \(0\ne1\). Subtraction means addition of an inverse: \(a-b=a+(-b)\). Division by nonzero \(b\) means multiplication by \(b^{-1}\): \(a/b=ab^{-1}\). Consequently, division by zero is not a difficult value that a calculator has failed to find; it is not defined by the field operations.

Many school rules are theorems derived from these axioms. Deriving them once makes their hypotheses visible.

Example 2: derive \(0a=0\) and \((-1)a=-a\)

Because \(0+0=0\), distributivity gives

[ 0a=(0+0)a=0a+0a. ]

Add the additive inverse of \(0a\) to both sides. The right side becomes \(0a\) and the left becomes \(0\), so \(0a=0\).

Next,

[ a+(-1)a=1a+(-1)a=(1+(-1))a=0a=0. ]

Thus \((-1)a\) is the additive inverse of \(a\), so uniqueness of additive inverses gives \((-1)a=-a\). No sign rule was assumed; it followed from field laws.

Two important cancellation theorems now follow:

  • additive cancellation: \(a+c=b+c\Rightarrow a=b\);
  • multiplicative cancellation: if \(c\ne0\), then \(ac=bc\Rightarrow a=b\).

Additive cancellation has no exception because every \(c\) has an additive inverse. Multiplicative cancellation requires \(c\ne0\) because zero has no multiplicative inverse.

Guided exercise: prove cancellation

Prove that \(ac=bc\) and \(c\ne0\) imply \(a=b\).

Use the inverse \(c^{-1}\) and preserve the grouping explicitly.

Starting from \(ac=bc\), multiply both sides by \(c^{-1}\):

[ (ac)c{-1}=(bc)c{-1}. ]

Associativity gives \(a(cc^{-1})=b(cc^{-1})\). Since \(c\ne0\), its inverse exists, so \(cc^{-1}=1\). Hence \(a=b\). The proof fails at exactly the place where \(c=0\): the symbol \(c^{-1}\) would not denote a field element.

3. Equations, identities, and domains

An expression such as \((x^2-1)/(x-1)\) has a domain. An equation asks for assignments in a stated universe that make two expressions equal. An identity is an equality true for every assignment in a specified common domain. These distinctions matter:

[ =x+1 ]

is true whenever \(x\ne1\), but the left expression is undefined at \(x=1\). Therefore the two expressions agree on \(\mathbb R\setminus\{1\}\) but do not define the same real-valued function if the right side is given domain \(\mathbb R\).

Before transforming an equation, record its original admissible set: the values for which every denominator is nonzero, every even root has a nonnegative radicand, every real logarithm has a positive argument, and every other operation is defined. Later algebra produces candidates; only candidates in that original set can be solutions.

Example 3: cancellation with a parameter

Solve over \(\mathbb R\) in \(x\):

[ a(x-2)=a(x+1), ]

where \(a\) is a real parameter.

A tempting step is to cancel \(a\), obtaining \(x-2=x+1\), hence \(-2=1\) and no solution. That conclusion assumes \(a\ne0\). Split the cases first.

  • If \(a\ne0\), multiplicative cancellation is legal and gives the contradiction \(-2=1\). There is no real \(x\).
  • If \(a=0\), the original equation is \(0=0\), true for every real \(x\).

Thus the solution set is

[ S_a= \[\begin{cases} \mathbb R,&a=0,\\ \varnothing,&a\ne0. \end{cases}\]

]

The exceptional parameter did not create a small correction; it changed no solutions into infinitely many.

Example 4: equal formulas, unequal domains

Factor and simplify

[ f(x)=. ]

The original domain is \(x\ne1\). Factoring gives

[ =x+1(x). ]

The notation \(x\ne1\) must survive cancellation. The graph of \(f\) is the line \(y=x+1\) with a hole at \((1,2)\). If we define \(g(x)=x+1\) on all real numbers, then \(f\) and \(g\) agree wherever \(f\) is defined, but \(f\ne g\) as functions because their domains differ.

A correct final statement is

[ f(x)=x+1xR{1}. ]

Domain checkpoint

Are \((x^2-4)/(x-2)\) and \(x+2\) the same function from \(\mathbb R\) to \(\mathbb R\)?

No. The rational expression is undefined at \(x=2\), while \(x+2\) is defined there. They have equal values on \(\mathbb R\setminus\{2\}\). Restricting both domains to that set would make them equal as functions.

4. Division, fractions, and clearing denominators

A fraction is multiplication by an inverse. This viewpoint explains the valid laws, assuming all displayed denominators are nonzero:

[ ac+bc=, abcd=, =. ]

It also exposes false distribution. In general,

[ ab+ac. ]

For the counterexample \(a=1,b=c=1\), the left side is \(1/2\) and the right side is \(2\). A fraction bar groups its entire numerator and denominator.

Clearing denominators means multiplying every term of an equation by a common denominator. On the original domain, that multiplier is nonzero, so the operation is reversible. If the excluded values are forgotten, the polynomial equation obtained afterward may admit values that never belonged to the rational equation.

Example 5: a rational equation whose only candidate is excluded

Solve over \(\mathbb R\):

[ =. ]

Step 1: state the domain. Since \(x^2-1=(x-1)(x+1)\),

[ x,qquad x. ]

Step 2: multiply by the nonzero common denominator. On that domain, \((x-1)(x+1)\ne0\), so

[ \[\begin{aligned} \frac1{x-1}=\frac2{(x-1)(x+1)} &\Longleftrightarrow x+1=2\\ &\Longleftrightarrow x=1. \end{aligned}\]

]

Step 3: enforce the original domain. The only algebraic candidate \(x=1\) is excluded. Therefore the solution set is \(\varnothing\).

Substitution into the original equation does not yield “false”; it yields an undefined expression. That is precisely why domain comes before manipulation.

Guided exercise: preserve an excluded value

Solve \(\dfrac{x}{x-3}=\dfrac{3}{x-3}+2\) over \(\mathbb R\).

Record \(x\ne3\), then multiply every term—not just selected numerators—by \(x-3\).

The original domain is \(x\ne3\). Multiplication by \(x-3\) gives

[ x=3+2(x-3)=2x-3. ]

Thus \(x=3\), but that candidate is outside the original domain. The solution set is empty. If the domain line is omitted, the final answer looks convincing but is invalid.

5. Products, zero, and factor reasoning

In \(\mathbb R\) and \(\mathbb C\), the zero-product property states

[ ab=0a=0b=0. ]

To prove it, suppose \(ab=0\). If \(a=0\), one branch is complete. If \(a\ne0\), multiply by \(a^{-1}\) to obtain \(b=0\). The converse also holds because \(0b=a0=0\); hence

[ ab=0a=0b=0. ]

This property is not merely about notation. It says the field has no nonzero zero divisors. It permits factoring to turn one equation into several branches. It does not permit concluding from \(ab=6\) that \(a=6\) or \(b=6\).

Example 6: solve by a reversible factor argument

Solve \(2x^3-5x^2-3x=0\) over \(\mathbb R\).

Factor before dividing by \(x\):

[ \[\begin{aligned} 2x^3-5x^2-3x&=0\\ x(2x^2-5x-3)&=0\\ x(2x+1)(x-3)&=0. \end{aligned}\]

]

By the zero-product property,

[ x=0,qquad 2x+1=0,qquadx-3=0. ]

Therefore \(S=\{-\tfrac12,0,3\}\). Substitution verifies every value.

Had we divided the original equation by \(x\), we would have assumed \(x\ne0\) and lost the solution \(x=0\). Factoring preserves the zero branch; premature division erases it.

Zero-product checkpoint

Why does \((x-4)(x+7)=0\) split into two equations, whereas \((x-4)(x+7)=10\) does not?

A product in a field equals zero only if at least one factor is zero. A nonzero product has many possible factor pairs, so neither factor is forced to equal \(10\).

6. Powers and radicals: laws with hypotheses

Integer exponent laws follow from repeated multiplication. When the displayed expressions are defined,

[ a^m an=a{m+n},(am)n=a^{mn}, (ab)n=an b^n. ]

For negative exponents, \(a^{-n}=1/a^n\) requires \(a\ne0\). Rational exponents bring root conditions. Over the reals, \(a^{1/n}\) is defined for every real \(a\) when \(n\) is odd, but only for \(a\ge0\) when \(n\) is even under the principal-root convention.

Powers generally do not distribute across addition:

[ (a+b)2=a2+2ab+b^2a2+b2. ]

Roots generally do not distribute across addition either: \(\sqrt{a+b}\ne\sqrt a+\sqrt b\). For \(a=b=1\), the sides are \(\sqrt2\) and \(2\). Even the product rule \(\sqrt{ab}=\sqrt a\sqrt b\) requires \(a,b\ge0\) over \(\mathbb R\). With \(a=b=-1\), the left side is \(1\) while the real square roots on the right are undefined; in principal complex arithmetic, the right side is \(i^2=-1\), still not \(1\).

Example 7: locate the earliest false power step

A proposed simplification reads

[ (x-2)2=x2-4x+4=x^2+4. ]

The first equality is valid by distributivity. The second drops the term \(-4x\) and is false except at \(x=0\). The repair is to retain all three terms.

A stronger audit uses a counterexample. At \(x=1\), the original value is \(1\), while \(x^2+4=5\). Because an identity is a universal claim, one input that makes the two sides unequal disproves it.

Example 8: solving after squaring requires a check

Solve \(\sqrt{x+2}=x\) over \(\mathbb R\).

Domain and sign. The radical requires \(x\ge-2\), but the left side is nonnegative, so equality additionally requires \(x\ge0\).

Generate candidates. Squaring is now safe as a necessary step:

[ x+2=x^2x^2-x-2=0 (x-2)(x+1)=0. ]

The candidates are \(2\) and \(-1\).

Check the original equation. At \(x=2\), \(\sqrt4=2\). At \(x=-1\), \(\sqrt1=1\ne-1\). Therefore \(S=\{2\}\).

The extraneous value did not appear because the arithmetic was poor. It appeared because \(t\mapsto t^2\) identifies \(t\) and \(-t\).

Guided exercise: a radical with two possible candidates

Solve \(\sqrt{2x+3}=x+1\) over \(\mathbb R\).

The right side must be nonnegative. Square, solve the quadratic, and return to the original equation.

The equality requires \(x+1\ge0\), so \(x\ge-1\). Squaring gives

[ 2x+3=(x+1)2=x2+2x+1, ]

hence \(x^2-2=0\) and \(x=\pm\sqrt2\). The value \(-\sqrt2<-1\) violates the sign condition and fails the original equation. The value \(\sqrt2\) verifies because both sides are nonnegative and their squares agree. Thus \(S=\{\sqrt2\}\).

7. Order laws and inequality direction

The real numbers form an ordered field. Exactly one of \(a<b\), \(a=b\), or \(a>b\) holds. Order is transitive. Addition preserves order:

[ a<ba+c<b+c. ]

Multiplication depends on sign:

[ \[\begin{array}{ll} c>0:&a<b\Longleftrightarrow ac<bc,\\ c<0:&a<b\Longleftrightarrow ac>bc. \end{array}\]

]

When \(c=0\), both products equal zero, so strict order is destroyed. Dividing an inequality by a symbolic expression is therefore a case split, not one routine move. The sign of the divisor decides both whether division is allowed and whether the order reverses.

Squaring also depends on location. On \([0,\infty)\), squaring is strictly increasing; on \((-\infty,0]\), it reverses order; across zero, comparing squares amounts to comparing absolute values.

Example 9: solve an inequality with a parameter multiplier

Solve \(a(x-1)>2a\) for real \(x\), classified by real \(a\).

First rewrite as \(a(x-3)>0\). Now split by the sign of \(a\).

  • If \(a>0\), division preserves order: \(x-3>0\), so \(x>3\).
  • If \(a<0\), division reverses order: \(x-3<0\), so \(x<3\).
  • If \(a=0\), the inequality becomes \(0>0\), which is false; there are no solutions.

Thus

[ S_a= \[\begin{cases} (3,\infty),&a>0,\\ \varnothing,&a=0,\\ (-\infty,3),&a<0. \end{cases}\]

]

Writing merely “divide by \(a\)” would miss both the reversal and the zero case.

Order checkpoint

A student has \(-3x<12\) and writes \(x<-4\). Identify and repair the error.

Division by the negative number \(-3\) reverses the strict inequality. The correct solution is \(x>-4\). Checking \(x=0\) confirms it: \(0<12\) is true.

8. Equivalent transformations and implication-only transformations

Suppose an original condition is \(P\) and the next line is \(Q\).

  • Write \(P\Longleftrightarrow Q\) when each implies the other on the stated domain. Solving either produces exactly the same solution set.
  • Write \(P\Longrightarrow Q\) when every original solution satisfies the new condition but the new condition may have extra solutions. Then \(Q\) produces candidates that must be checked in \(P\).
  • A transformation can also be invalid: neither direction may hold.

For equations over a field, these operations are equivalences:

  • add or subtract the same defined expression on both sides;
  • multiply or divide both sides by an expression known to be nonzero;
  • apply a one-to-one function on a domain containing both sides.

These commonly give only implications unless extra conditions are supplied:

  • square both sides;
  • raise both sides to an even power;
  • multiply by an expression that might be zero when reversing the step;
  • replace \(\sqrt{u^2}\) by \(u\) without knowing \(u\ge0\);
  • apply a periodic function such as sine.

Example 10: label a solution chain by direction

Audit the equation \(x-1=\sqrt{x+5}\) over \(\mathbb R\).

The equality forces \(x-1\ge0\), so \(x\ge1\). Then

[ \[\begin{aligned} x-1=\sqrt{x+5} &\Longrightarrow (x-1)^2=x+5\\ &\Longleftrightarrow x^2-3x-4=0\\ &\Longleftrightarrow (x-4)(x+1)=0. \end{aligned}\]

]

The first arrow is only an implication before sign information is incorporated. The candidates are \(x=4,-1\). The condition \(x\ge1\) eliminates \(-1\); substitution confirms \(4-1=\sqrt9=3\). Hence \(S=\{4\}\).

Alternatively, after explicitly restricting to \(x\ge1\), both sides are nonnegative and squaring becomes reversible. Good writing makes that restricted universe visible.

Example 11: multiplication can hide a false converse

Compare

[ =1 x-2=x-2. ]

On the original domain \(x\ne2\), multiplying by \(x-2\) is reversible and the two conditions are equivalent. On all reals, the second equation is true at \(x=2\) while the first is undefined. Therefore the unrestricted converse is false.

A precise chain is

[ =1 x-2=x-2 x. ]

The condition is part of the equivalence, not a footnote.

Guided exercise: choose the arrow

For real \(x\), insert \(\Longleftrightarrow\) or \(\Longrightarrow\):

[ x=5?(x-5)(x+2)=0. ]

Then state the condition under which the implication can become an equivalence.

The correct unrestricted relation is

[ x=5(x-5)(x+2)=0. ]

The right side also holds at \(x=-2\). Under the additional condition \(x\ne-2\), the zero-product property forces \(x-5=0\), so the statements become equivalent.

9. Conditional identities and substitution

An identity is always relative to a domain. The equation

[ (a-b)(a+b)=a2-b2 ]

is an unconditional polynomial identity over every commutative field. In contrast,

[ =a+b ]

is conditional: it requires \(a\ne b\). The factored numerator does not repair the zero denominator in the original expression.

Substitution must also respect domains. If \(u=v\) and a formula contains \(1/u\), replacing it by \(1/v\) is valid only where both denominators are defined. If \(u=v\) as values at one point, that does not prove the formulas are identical for every point. If two functions agree on one domain but one has a larger natural domain, substitution into a statement about function domains can change the claim.

Example 12: state a conditional identity completely

Simplify

[ . ]

Factor the difference of squares:

[ =x+a, ]

provided \(x-a\ne0\). The complete identity is

[ =x+axa. ]

If \(a\) is fixed, the left side has a removable discontinuity at \(x=a\); its limiting value there is \(2a\), but its actual value remains undefined. A limit or continuous extension is not the same object as the original formula.

Identity checkpoint

Complete the statement:

[ =u2+uv+v2. ]

The condition is \(u\ne v\). Factoring proves the equality on that domain. When \(u=v\), the original quotient is undefined even though the polynomial on the right is defined.

10. Parameter equations: classify before canceling

A parameter is held fixed while the principal variable varies. A complete answer states how the solution set changes with the parameter. The robust workflow is:

  1. state restrictions inherited from denominators, radicals, or logarithms;
  2. collect the coefficient of the principal variable;
  3. identify parameter values that make a planned divisor zero;
  4. solve the generic nonzero case;
  5. substitute every exceptional parameter into the original equation; and
  6. distinguish no solution, one or finitely many solutions, and infinitely many solutions.

Example 13: one equation, three parameter behaviors

Solve over \(\mathbb R\) in \(x\):

[ (a-2)x=a^2-4. ]

Factor the right side:

[ (a-2)x=(a-2)(a+2). ]

  • If \(a\ne2\), divide by \(a-2\) to obtain the unique solution \(x=a+2\).
  • If \(a=2\), return to the original equation: \(0\cdot x=0\). Every real \(x\) is a solution.

Thus

[ S_a= \[\begin{cases} \{a+2\},&a\ne2,\\ \mathbb R,&a=2. \end{cases}\]

]

For comparison, changing the right side to \(a+2\) would produce no solution at \(a=2\), because the exceptional equation would be \(0x=4\). Exceptional values must be substituted, not guessed from the generic formula.

Guided exercise: classify all cases

Solve \((k-1)x=k-1\) over \(\mathbb R\).

The planned divisor is \(k-1\). Separate the case in which it vanishes.

If \(k\ne1\), divide by \(k-1\) and obtain \(x=1\). If \(k=1\), the original equation is \(0=0\), so every real \(x\) works. Therefore

[ S_k= \[\begin{cases}\{1\},&k\ne1,\\\mathbb R,&k=1.\end{cases}\]

]

11. Common mistakes

Mistake 1: cancel terms across addition

From \((x+3)/x\), one cannot “cancel the \(x\)” to get \(3\). Cancellation removes a common factor, not a term. A valid rewrite is

[ =+3x=1+3x(x). ]

At \(x=1\), the original value is \(4\), immediately refuting the proposed value \(3\).

Mistake 2: divide by the unknown before checking zero

From \(x(x-5)=0\), division by \(x\) produces \(x=5\) but assumes \(x\ne0\) and loses the solution \(0\). Use the zero-product property or split into \(x=0\) and \(x\ne0\).

Mistake 3: distribute a power or root over a sum

Neither \((a+b)^n=a^n+b^n\) nor \(\sqrt{a+b}=\sqrt a+\sqrt b\) is a general real law. Expand powers by multiplication; test radical claims with small nonnegative counterexamples.

Mistake 4: clear denominators without recording restrictions

Multiplying by a common denominator is reversible only on the set where that denominator is nonzero. Write exclusions before multiplying and filter the final candidate set through them.

Mistake 5: reverse an inequality automatically

Addition never reverses order. Multiplication or division reverses it exactly when the multiplier is negative. For a symbolic multiplier, determine its sign or split into cases.

Mistake 6: treat an implication as an equivalence

Squaring, even powers, and some trigonometric or absolute-value operations merge values. Use \(\Rightarrow\) while generating candidates, then verify in the original condition. If sign restrictions make the operation one-to-one, say so and then use \(\Leftrightarrow\).

Mistake 7: believe software has supplied missing assumptions

A computer algebra system may simplify a generic expression, return candidates over the complex numbers, or retain a piecewise result. Its output is evidence to interpret. It does not retroactively specify the problem’s universe or prove that every transformation preserved the domain.

Earliest-invalid-step audit

A student solves

[ \[\begin{aligned} \frac{x+1}{x-2}&=1,\\ x+1&=x-2,\\ 1&=-2, \end{aligned}\]

]

and concludes there is no solution. Is any displayed transformation invalid?

The conclusion is correct, and the transformation is valid after stating the original restriction \(x\ne2\). On that domain, multiplication by \(x-2\) is reversible. The contradiction \(1=-2\) shows no admissible \(x\) exists. The audit finds an omitted justification, not a false algebraic line.

12. Examination strategy clinic

Hard questions often combine a parameter, a nonreversible operation, and a boundary at which the number of solutions changes. The task is not merely to solve a quadratic; it is to decide which roots survive the original equation.

Example 14: classify a parameterized radical equation

For real \(a\), determine the number and values of real solutions of

[ =x-a. ]

Step 1: isolate the sign information

The square root is nonnegative, so every solution must satisfy

[ x-a. ]

Set \(y=x-a\). Then \(y\ge0\) and \(x=y+a\). The radicand becomes

[ x+a=y+2a. ]

The equation is

[ =y,qquad y. ]

Step 2: square and solve the candidate equation

Because the right side is known to be nonnegative, squaring will be reversible for candidates that satisfy the resulting equation and \(y\ge0\):

[ y+2a=y^2 y^2-y-2a=0. ]

The discriminant is \(\Delta=1+8a\), and

[ y_{}=2. ]

Real candidates require \(a\ge-1/8\).

Step 3: classify by the sign condition \(y\ge0\)

  • If \(a<-1/8\), the quadratic has no real roots, so there are no solutions.
  • If \(a=-1/8\), the double root is \(y=1/2\), giving \(x=y+a=3/8\).
  • If \(-1/8<a<0\), then \(0<\sqrt{1+8a}<1\), so both \(y_+\) and \(y_-\) are positive. Both give solutions.
  • If \(a=0\), the roots are \(y=1\) and \(y=0\); both are nonnegative, so both survive.
  • If \(a>0\), then \(\sqrt{1+8a}>1\), so \(y_-<0\) and must be rejected, while \(y_+>0\) survives.

Converting back to \(x=a+y\) gives

[ S_a= \[\begin{cases} \varnothing,&a<-\frac18,\\[2mm] \left\{\frac38\right\},&a=-\frac18,\\[2mm] \left\{a+\frac{1-\sqrt{1+8a}}2, a+\frac{1+\sqrt{1+8a}}2\right\}, &-\frac18<a\le0,\\[2mm] \left\{a+\frac{1+\sqrt{1+8a}}2\right\},&a>0. \end{cases}\]

]

Step 4: prove that no hidden extraneous root remains

For every retained root, \(y\ge0\) and the quadratic equation gives \(y+2a=y^2\). Hence

[ ==|y|=y. ]

So every retained candidate satisfies the original equation. This last argument is stronger than checking a few numerical parameters: it verifies the whole classified family.

Examination audit

A full-credit solution must identify both transition parameters. The value \(a=-1/8\) changes the discriminant, while \(a=0\) changes the sign of the smaller root. Reporting only the quadratic formula misses the problem’s central reasoning.

Clinic transfer

For real \(p\), classify the real solutions of \((p-1)(x+2)=p^2-1\).

Factor \(p^2-1=(p-1)(p+1)\). If \(p\ne1\), cancellation gives \(x+2=p+1\), so \(x=p-1\). If \(p=1\), the original equation is \(0=0\), so every real \(x\) is a solution. Therefore

[ S_p= \[\begin{cases}\{p-1\},&p\ne1,\\\mathbb R,&p=1.\end{cases}\]

]

The apparent special value \(p=-1\) causes no loss: the divisor \(p-1\) is then nonzero and the unique solution is \(x=-2\).

13. Exercises by purpose and difficulty

Attempt the problems without software. For an equation, state the universe and restrictions before transforming it. For an audit problem, identify the earliest line that is not justified. Concise answers follow all problems; selected Level C and D solutions then model full examination writing.

Level A — laws and fluency

  1. Name the principal law used in \(7(x+y)=7x+7y\).
  1. Simplify \(4(3x-2)-5(x+1)\) and name the laws that justify regrouping.
  1. State the excluded values for \(\dfrac{x+4}{x^2-9}\).
  1. Complete the identity \(\dfrac{x^2-16}{x-4}=\underline{\hspace{1cm}}\) with its condition.
  1. Solve \((2x-1)(x+5)=0\) over \(\mathbb R\).
  1. Solve \(-4x\ge20\) and explain the direction of the inequality.
  1. Give a counterexample to \(\sqrt{u+v}=\sqrt u+\sqrt v\) with \(u,v\ge0\).
  1. Decide whether \(x=3\Longrightarrow x^2=9\) and its converse are true.

Level B — connected reasoning

  1. Prove additive cancellation from the field laws.
  1. Solve \(\dfrac{x+2}{x-1}=3\) over \(\mathbb R\) and verify the result.
  1. Solve \(x(x+4)=0\) without dividing by \(x\). Explain what division would lose.
  1. Determine the common domain on which \(\dfrac{x^2-x}{x}\) and \(x-1\) agree.
  1. Solve \(\sqrt{x+6}=x\) over \(\mathbb R\).
  1. For \(c>0\), solve \(c(2x-1)<3c\). Then repeat for \(c<0\).
  1. Disprove \((a/b)+(c/d)=(a+c)/(b+d)\) as a universal fraction law.
  1. State precisely when \(\dfrac{r^3-s^3}{r-s}=r^2+rs+s^2\).

Level C — synthesis and audits

  1. Solve \(\dfrac{2}{x-2}=\dfrac{x}{x-2}-1\) and explain the status of every candidate.
  1. A student writes \(\sqrt{(x-3)^2}=x-3\). Find the largest real domain on which this identity is valid and give a counterexample outside it.
  1. Classify the solutions of \((m+2)x=m^2-4\) by real \(m\).
  1. Solve \(\sqrt{3x+4}=x+2\) over \(\mathbb R\), labeling implication-only steps.
  1. For real \(a\), solve \(a(x+1)\le0\) in \(x\).
  1. Find the earliest invalid step:

[ =0 x^2-4=0 x=. ]

  1. Determine whether \(\dfrac1{a+b}=\dfrac1a+\dfrac1b\) can ever hold for nonzero real \(a,b,a+b\). Describe all such pairs.
  1. Solve \(|x-1|=x+3\) and state where squaring is reversible.

Level D — proof, construction, and classification

  1. Prove from field laws that \((-a)(-b)=ab\).
  1. Construct an equation for which squaring produces exactly two extraneous real roots and justify your construction.
  1. For real \(a\), classify the solutions of \(\dfrac{x-a}{x-1}=0\).
  1. For real \(k\), classify the real solutions of \((k-2)x=kx+1\).
  1. Prove that if \(0<a<b\), then \(1/a>1/b\), using only ordered-field operations and explicit sign information.
  1. Determine every real \(a\) for which \(\sqrt{x+a}=x\) has exactly two real solutions.
  1. Let \(u,v\in\mathbb R\). Determine necessary and sufficient conditions for \(\sqrt{u^2v^2}=uv\).
  1. A proposed proof starts with \(p=q\ne0\), multiplies by \(p\), subtracts \(q^2\), factors, divides by \(p-q\), and concludes \(2=1\). Reconstruct the lines and identify exactly why the division is invalid.

Concise answers

  1. Distributivity.
  2. \(12x-8-5x-5=7x-13\); distributivity, then associativity and commutativity.
  3. \(x\ne-3,3\).
  4. \(x+4\), provided \(x\ne4\).
  5. \(x=1/2\) or \(x=-5\).
  6. \(x\le-5\); division by \(-4\) reverses order.
  7. \(u=v=1\): \(\sqrt2\ne2\).
  8. The implication is true; the converse is false because \(x=-3\) also has square \(9\).
  9. Add \(-c\) to both sides of \(a+c=b+c\) and use associativity and additive inverses.
  10. Domain \(x\ne1\); \(x+2=3x-3\), so \(x=5/2\), which verifies.
  11. \(x=0\) or \(x=-4\); division by \(x\) would discard \(0\).
  12. \(\mathbb R\setminus\{0\}\).
  13. Sign requires \(x\ge0\). Squaring gives \(x^2-x-6=0\), so candidates \(3,-2\); only \(3\) verifies.
  14. If \(c>0\), \(x<2\); if \(c<0\), \(x>2\).
  15. Take \(a=b=c=d=1\): the left side is \(2\), the right side \(1\).
  16. It holds whenever \(r\ne s\); that is the original quotient’s domain.
  17. Domain \(x\ne2\). Clearing gives \(2=x-(x-2)=2\), so every admissible value works: \(\mathbb R\setminus\{2\}\).
  18. Since \(\sqrt{(x-3)^2}=|x-3|\), equality with \(x-3\) holds exactly for \(x\ge3\). At \(x=2\), the sides are \(1\) and \(-1\).
  19. If \(m\ne-2\), \(x=m-2\); if \(m=-2\), every real \(x\) solves \(0=0\).
  20. The radical requires \(x\ge-4/3\); on that domain the right side is already positive. Squaring gives \(3x+4=(x+2)^2\), hence \(x(x+1)=0\); both \(0\) and \(-1\) lie in the domain and verify.
  21. If \(a>0\), \(x\le-1\); if \(a<0\), \(x\ge-1\); if \(a=0\), all real \(x\).
  22. The first equivalence omits \(x\ne2\). The numerator equation produces \(x=2\), which is excluded; the actual solution is \(x=-2\).
  23. Multiplication by \(ab(a+b)\) gives \(ab=(a+b)^2\). Equivalently \(a^2+ab+b^2=0\). Its discriminant in \(a/b\) is \(-3\), so no admissible real pairs exist.
  24. The right side must be nonnegative, which holds for \(x\ge-3\). Squaring gives \((x-1)^2=(x+3)^2\), hence \(x=-1\); it verifies. On \(x\ge-3\), both sides are nonnegative, so squaring is reversible.
  25. See the full solution below.
  26. One answer is \(\sqrt{x^2}= -1\). It has no real solution, while squaring gives \(x^2=1\) with two extraneous candidates \(\pm1\).
  27. Domain \(x\ne1\). A fraction is zero exactly when its numerator is zero, so \(x=a\) if \(a\ne1\); if \(a=1\), the candidate is excluded and there is no solution.
  28. Rearrangement gives \(-2x=1\), independent of \(k\), so \(x=-1/2\) for every real \(k\).
  29. See the full solution below.
  30. Squaring \(\sqrt{x+a}=x\) under \(x\ge0\) gives \(x^2-x-a=0\). It has two distinct nonnegative roots when \(-1/4<a<0\) (both roots are positive) and also at \(a=0\) (the roots are \(0\) and \(1\)). Thus the complete parameter set is \(-1/4<a\le0\).
  31. The left side is \(|uv|\), so equality holds exactly when \(uv\ge0\).
  32. See the full solution below.

Selected full solutions

First, \((-a)b\) is the additive inverse of \(ab\) because

[ ab+(-a)b=(a+(-a))b=0b=0. ]

Hence \((-a)b=-(ab)\). Apply the same fact with \(b\) replaced by \(-b\): \((-a)(-b)=-[a(-b)]\). But \(a(-b)=-(ab)\), so

[ (-a)(-b)=-[-(ab)]=ab. ]

The final equality uses uniqueness of additive inverses, or the previously derived rule \(-(-t)=t\).

From \(0<a<b\), both \(a\) and \(b\) are positive, so \(ab>0\) and division by \(ab\) preserves order. Divide \(a<b\) by \(ab\):

[ <. ]

Simplifying with nonzero \(a,b\) gives \(1/b<1/a\), equivalently \(1/a>1/b\). Every cancellation is licensed by positivity, which implies nonzero.

Starting from \(p=q\ne0\):

[ \[\begin{aligned} p&=q,\\ p^2&=pq,\\ p^2-q^2&=pq-q^2,\\ (p-q)(p+q)&=q(p-q). \end{aligned}\]

]

The false proof divides by \(p-q\) to obtain \(p+q=q\), then substitutes \(p=q\) to claim \(2q=q\) and \(2=1\). But the premise \(p=q\) makes \(p-q=0\). The cancellation step divides by zero and is undefined. Every earlier line is valid.

14. Cumulative retrieval

These problems combine the language of Chapter 1.1 and the number-system conditions of Chapter 1.2. The method is deliberately not named in the prompt.

Quantify cancellation correctly

Write a quantified real-number statement expressing multiplicative cancellation, then negate it. Explain why deleting the nonzero condition makes the original claim false.

A precise statement is

[ (a,b,cR) [(ac=bcc)a=b]. ]

Its negation is

[ (a,b,cR) [ac=bccab]. ]

No such triple exists. Without \(c\ne0\), choose \(a=1,b=2,c=0\); then \(ac=bc=0\) although \(a\ne b\).

Domain, density, and identity

Let

[ f(x)=. ]

State its real domain, simplify it there, and decide whether rational inputs are dense in its domain. Is the simplified polynomial the same real function?

The denominator excludes \(x=\sqrt2\), so the domain is \(\mathbb R\setminus\{\sqrt2\}\). Factoring over the reals gives

[ f(x)=x+(x). ]

The rationals remain dense in this punctured domain: every open interval around an admissible point contains rationals, and the removed point is irrational in any case. The polynomial \(g(x)=x+\sqrt2\) with natural domain \(\mathbb R\) is not the same function as \(f\) because \(g(\sqrt2)=2\sqrt2\) while \(f(\sqrt2)\) is undefined.

15. Topic checkpoint

First answer the static challenge: for which real \(a\) may one cancel \(a-3\) from

[ (a-3)(x+1)=(a-3)(2x-5)? ]

If \(a\ne3\), cancellation is valid and leads to \(x=6\). If \(a=3\), the original equation is \(0=0\) and every real \(x\) is a solution. The interactive checkpoint below varies the type of transformation; explain the domain and direction before selecting an answer.

16. Summary and next step

  • Equality permits substitution, but reversing an applied operation requires that operation to be one-to-one on the relevant domain.
  • Field laws justify expansion, regrouping, inverses, and cancellation. Multiplicative cancellation requires a nonzero factor.
  • An expression’s original domain survives simplification. Equal formulas need not define equal functions when their domains differ.
  • Clearing denominators is reversible on the original domain; excluded values never become solutions later.
  • The zero-product property creates branches. Dividing by an unknown factor can destroy the zero branch.
  • Power, radical, and fraction laws have hypotheses. One decisive counterexample refutes a proposed universal identity.
  • Multiplication or division of an inequality preserves direction for a positive factor, reverses it for a negative factor, and destroys strict order for a zero factor.
  • Equivalent steps preserve the solution set. Implication-only steps generate candidates that require verification in the original condition.
  • Parameter problems demand a generic case and explicit exceptional cases.
  • A strong solution records universe, domain, direction, exceptions, and a final check.

Continue to Proof methods for algebra, which organizes these valid local steps into complete direct, contrapositive, contradiction, case, equivalence, existence, uniqueness, counterexample, and induction arguments.

References and further study

  • OpenStax, College Algebra 2e provides an openly licensed review of real-number properties, equations, rational expressions, radicals, and inequalities. This chapter’s exposition and original problems were independently written; the text is a supporting reference rather than a source of copied exercises.
  • Richard Hammack, Book of Proof develops logic, direct proof, contrapositive, contradiction, and careful use of definitions; it is especially useful before the next chapter.
  • SymPy simplification documentation explains symbolic expansion, factoring, cancellation, and simplification. Compare its transformations with the domain audit in this chapter.
  • SymPy equation-solving guidance documents exact solvers and the interpretation of returned candidates.

Using this lesson with fcmath and SymPy

Use exact parsing and symbolic subtraction to check a polynomial identity. A zero result supports equivalence on the common domain; it does not compare function domains for you.

import sympy as sp

from fcmath import parse_expression

x, a = sp.symbols("x a", real=True)
original = parse_expression("3*(2*x - 5) - 2*(x + 4)", variables=(x,))
simplified = parse_expression("4*x - 23", variables=(x,))
sp.simplify(original - simplified)

\(\displaystyle 0\)

Auditing a conditional identity

SymPy can cancel the common factor in a rational expression. Record the lost input separately:

rational = (x**2 - a**2) / (x - a)
cancelled = sp.cancel(rational)
excluded = sp.solve(sp.denom(rational), x)
cancelled, excluded
(a + x, [a])

The output \(a+x\) is valid where \(x\ne a\). The exclusion list is part of the mathematical answer.

Solving an equation and checking the original

For \(\sqrt{x+2}=x\), the manual method first established \(x\ge0\), squared to generate candidates, and checked both candidates. The equivalent SymPy workflow keeps those stages separate:

candidate_equation = sp.Eq(x + 2, x**2)
candidates = sp.solve(candidate_equation, x)
valid = [
    value
    for value in candidates
    if value >= 0 and sp.simplify(sp.sqrt(value + 2) - value) == 0
]
candidates, valid
([-1, 2], [2])

The candidate list is \([-1,2]\) and the validated list is \([2]\). Asking a solver for candidates does not remove the need to inspect the original domain.

Checking a parameter classification

The examination-clinic substitution \(y=x-a\) reduces the radical equation to a quadratic. Its discriminant and roots expose the transition values:

y = sp.symbols("y", real=True)
polynomial = y**2 - y - 2 * a
discriminant = sp.discriminant(polynomial, y)
roots = sp.solve(polynomial, y)
discriminant, roots
(8*a + 1, [1/2 - sqrt(8*a + 1)/2, sqrt(8*a + 1)/2 + 1/2])

The symbolic output gives \(1+8a\) and \((1\pm\sqrt{1+8a})/2\). Human sign analysis of the smaller root supplies the second boundary \(a=0\).

Solving an inequality with an explicit real domain

inequality_solution = sp.solve_univariate_inequality(-3 * x < 12, x)
inequality_solution

\(\displaystyle -4 < x\)

SymPy returns \(-4<x\). Read the relation as a real solution set and connect it to the ordered-field rule: division by the negative coefficient reverses the inequality.

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