How to Read and Write an Algebraic Argument

Learn to distinguish claims, preserve logical direction, justify each algebraic step, and repair invalid solutions.

Why this matters

A correct final number does not by itself make a mathematical solution correct. A solution is an argument: it begins with stated objects and conditions, passes through justified steps, and ends with a conclusion that answers the question. That standard matters in advanced courses because an unnoticed lost case, illegal cancellation, or reversed implication can survive many lines of otherwise flawless algebra.

This chapter teaches a method you can use on every later page:

  1. identify what each line claims;
  2. record the domain and assumptions;
  3. decide whether a step preserves equivalence or only implication;
  4. name the law or definition that permits the step;
  5. check the conclusion against the original problem.

The material may look elementary, but the habits are not. The most demanding algebra examinations often hide their difficulty in a condition, a quantifier, or an exceptional parameter rather than in long arithmetic.

What you should be able to do

After completing the chapter, you should be able to:

  • distinguish an expression, equation, identity, inequality, predicate, and solution set;
  • use \(=\), \(\approx\), \(\Rightarrow\), and \(\Leftrightarrow\) with correct meanings;
  • write aligned calculations in which equality is transitive from line to line;
  • state domain restrictions before simplifying or solving;
  • decide whether a transformation is reversible;
  • identify and explain the first invalid step in a proposed solution;
  • repair an invalid argument without concealing the original error;
  • present exact and approximate results appropriately;
  • write a concise proof or counterexample rather than relying on examples as proof.

Prerequisites

You need signed arithmetic, fraction arithmetic, order of operations, and the meaning of a variable. If the warm-up reveals a gap, continue to the next chapter’s targeted repair route before attempting the examination set.

Five-minute retrieval warm-up

Work without a calculator.

  1. Evaluate \(7-3(2-5)\).
  2. Compute \(\frac34-\frac56\) exactly.
  3. Solve \(5x-2=18\) and check the result.
  4. Evaluate \(2a^2-ab\) when \(a=-2\) and \(b=3\).
  5. Is \(0.333\) equal to \(1/3\), or is it an approximation?
  1. \(7-3(-3)=16\).
  2. \(\frac{9}{12}-\frac{10}{12}=-\frac1{12}\).
  3. \(5x=20\), so \(x=4\); indeed \(5(4)-2=18\).
  4. \(2(-2)^2-(-2)(3)=8+6=14\).
  5. \(0.333=333/1000\) exactly and is only an approximation to \(1/3\). The repeating decimal \(0.\overline{3}\) equals \(1/3\).

1. Mathematical objects and mathematical claims

An expression names a mathematical object but does not assert that something is true. For example,

\[ 3x^2-2x+1 \]

is an expression. It can be evaluated or transformed. It is not true or false.

An equation asserts that two expressions have equal values:

\[ 3x^2-2x+1=0. \]

Its truth may depend on \(x\). An inequality compares values, as in \(3x^2-2x+1>0\). A predicate is a statement containing variables that becomes true or false when the variables are specified. Thus \(P(x)\) might mean “\(x^2<4\).”

An identity is an equation true for every value in a stated common domain. For example,

\[ (x+1)^2=x^2+2x+1 \]

is an identity for all real or complex \(x\). In contrast, \(x^2=4\) is a conditional equation: it is true only for \(x=-2\) and \(x=2\) over the reals.

A solution set collects every allowed value that makes a predicate true. The solution set of \(x^2<4\) over the reals is \((-2,2)\). The phrase “over the reals” is part of the problem; over the integers the solution set would be \(\{-1,0,1\}\).

Example 1: classify before calculating

Classify each item.

  1. \(2x+5\)
  2. \(2x+5=11\)
  3. \(2(x+3)=2x+6\)
  4. \(x/0\)
  5. \(x^2\ge 0\) for every real \(x\)

Solution. Item 1 is an expression. Item 2 is a conditional equation with solution set \(\{3\}\). Item 3 is an identity over any usual number system in which the operations are defined. Item 4 is not a real or complex expression with a defined value because division by zero is undefined. Item 5 is a universally quantified true statement over the reals; its justification uses order, not merely several examples.

Checkpoint: object or claim?

Why is “simplify \(x^2-1\)” incomplete unless a desired form or purpose is given?

The expression already has a valid expanded form. It could instead be factored as \((x-1)(x+1)\), but neither representation is universally simpler. Expanded form exposes coefficients; factored form exposes zeros. A mathematical task should specify the information or operation sought.

2. Four symbols that must not be confused

Equality: \(=\)

The symbol \(=\) says that the expressions on its two sides denote the same value. It is symmetric and transitive: if \(A=B\), then \(B=A\); if \(A=B\) and \(B=C\), then \(A=C\).

An equality chain such as

\[ (x+2)^2=x^2+4x+4=(x^2+4x)+4 \]

claims that every displayed expression has the same value under the stated conditions.

Approximation: \(\approx\)

The symbol \(\approx\) reports closeness under an understood precision:

\[ \sqrt2\approx1.4142. \]

The right side is not equal to \(\sqrt2\). Good work states the precision or error when it matters, for example \(\sqrt2=1.41421356\ldots\) and \(\sqrt2\approx1.4142\) to four decimal places.

Implication: \(\Rightarrow\)

The statement \(P\Rightarrow Q\) means that whenever \(P\) is true, \(Q\) must be true. It does not say that \(Q\) forces \(P\).

\[ x=3\Rightarrow x^2=9 \]

is true, while its converse is false over the reals because \(x=-3\) also has square \(9\).

Equivalence: \(\Leftrightarrow\)

The statement \(P\Leftrightarrow Q\) means both \(P\Rightarrow Q\) and \(Q\Rightarrow P\). The statements have the same truth value under the stated conditions. In solving,

\[ 3x-5=7\Leftrightarrow3x=12\Leftrightarrow x=4 \]

because adding \(5\) and dividing by nonzero \(3\) are reversible.

Example 2: a one-way operation

Suppose \(x=-2\). Squaring gives \(x^2=4\), so

\[ x=-2\Rightarrow x^2=4. \]

But \(x^2=4\) allows both \(x=-2\) and \(x=2\). Therefore we may not replace the arrow with \(\Leftrightarrow\). Squaring an equation can enlarge its solution set. When we square during a solution, the outputs are candidates until checked in the original equation.

Example 3: when multiplication is reversible

For real \(a\), compare

\[ x=5 \quad\text{and}\quad ax=5a. \]

Multiplying by \(a\) always proves \(x=5\Rightarrow ax=5a\). The reverse direction requires \(a\ne0\) so that division by \(a\) is defined. If \(a=0\), the second equation is \(0=0\) and every real \(x\) satisfies it. Hence

\[ a\ne0\implies (x=5\Leftrightarrow ax=5a). \]

This is the prototype for parameter cases: never cancel a parameter before checking whether it can be zero.

3. Conditions and domains belong to the argument

The domain is the set of allowed inputs. It may be stated by the problem or forced by an expression. Over the real numbers:

  • a denominator cannot be zero;
  • an even-index radical needs a nonnegative radicand;
  • a logarithm needs a positive argument;
  • an inverse trigonometric function uses a specified principal range.

Restrictions must be recorded before an operation hides them.

Example 4: cancellation changes visible domain information

Consider

\[ R(x)=\frac{x^2-1}{x-1}. \]

The original denominator requires \(x\ne1\). Factoring gives

\[ R(x)=\frac{(x-1)(x+1)}{x-1}=x+1,\qquad x\ne1. \]

The formulas agree for every \(x\) in the original domain, but they do not define the same real-valued function if \(x+1\) is assigned domain \(\mathbb R\). The first function has a hole at \(x=1\); the second does not. The correct final line keeps the restriction.

Example 5: a radical candidate must be checked

Solve \(\sqrt{x+2}=x\) over the reals.

Conditions. The radical requires \(x\ge-2\), and because its value is nonnegative, equality requires \(x\ge0\).

Implication step. Squaring gives

\[ \sqrt{x+2}=x\Rightarrow x+2=x^2, \]

so \(x^2-x-2=0\) and \((x-2)(x+1)=0\). The candidates are \(2\) and \(-1\).

Check. At \(x=2\), \(\sqrt4=2\). At \(x=-1\), the left side is \(1\) but the right side is \(-1\). Therefore the solution set is \(\{2\}\). The invalid candidate was created by a nonreversible step; it is not an arithmetic accident.

4. How to write a readable algebraic derivation

A calculation should have one governing relation per line. Align equal signs or implication symbols and place reasons beside nontrivial changes. Do not use an equality sign to mean “the next thing I did.”

Example 6: repair a running calculation

The line

\[ 3(4)^2-5=3\cdot16=48-5=43 \]

is invalid because it claims \(3(4)^2-5=3\cdot16\), which omits \(-5\). A readable calculation is

\[ \begin{aligned} 3(4)^2-5 &=3\cdot16-5 &&\text{(evaluate the power)}\\ &=48-5 &&\text{(multiply)}\\ &=43. &&\text{(subtract)} \end{aligned} \]

Every adjacent pair is equal, so transitivity justifies the full chain.

Example 7: solving with reasons

Solve \(4(2x-3)+5=3x+14\).

\[ \begin{aligned} 4(2x-3)+5&=3x+14 &&\text{(given)}\\ 8x-12+5&=3x+14 &&\text{(distributive law)}\\ 8x-7&=3x+14 &&\text{(combine like terms)}\\ 5x-7&=14 &&\text{(subtract $3x$ from both sides)}\\ 5x&=21 &&\text{(add $7$ to both sides)}\\ x&=\frac{21}{5}. &&\text{(divide by nonzero $5$)} \end{aligned} \]

Substitution gives \(4(42/5-3)+5=133/5\) and \(3(21/5)+14=133/5\). The solution set is \(\{21/5\}\).

A compact line-audit protocol

For each arrow or equality, ask:

  • Object: What statement is being transformed?
  • Operation: What exactly was done?
  • Permission: Which law or condition permits it?
  • Direction: Does the new line imply the old one as well?
  • Residue: Did the operation introduce a restriction or candidate check?

This protocol is deliberately slow now. With practice it becomes an internal habit, leaving written annotations only where a reader genuinely needs them.

5. Definition, theorem, example, and counterexample

A definition fixes meaning. It is not proved. A theorem is a claim proved from definitions and accepted results. An example illustrates a definition or result. A counterexample refutes a universal claim by satisfying its hypothesis and violating its conclusion.

Examples can reveal a pattern but cannot establish a statement about every number. Counterexamples have the opposite asymmetry: one can be enough to disprove a universal statement.

Example 8: test a universal claim

Claim: for all real \(a,b\),

\[ \sqrt{a+b}=\sqrt a+\sqrt b. \]

Take \(a=b=1\). The left side is \(\sqrt2\) and the right side is \(2\), so they are not equal. This one counterexample disproves the universal claim. It also shows why checking small values is useful before attempting a proof.

The counterexample is valid because both sides are defined. Choosing \(a=-1\), \(b=0\) over the reals would not refute the claim cleanly; it would instead expose an unstated domain problem.

Example 9: examples are not a proof

The values \(1^3-1\), \(2^3-2\), and \(3^3-3\) are divisible by \(3\). That evidence suggests the statement “\(n^3-n\) is divisible by \(3\) for every integer \(n\),” but it does not prove it.

A proof uses the three possible integer remainders modulo \(3\). Here \(u\equiv v\pmod 3\) means that \(u\) and \(v\) have the same remainder after division by \(3\):

  • if \(n=3k\), then \(n^3-n\) is divisible by \(3\);
  • if \(n=3k+1\), then \(n^3-n\equiv1^3-1\equiv0\pmod3\);
  • if \(n=3k+2\), then \(n^3-n\equiv2^3-2=6\equiv0\pmod3\).

Every integer belongs to one case, so the argument is general.

6. Exact values, approximations, and checks

Use exact forms while they preserve structure. Fractions expose ratios; radicals expose algebraic origin; \(\pi\) exposes angle geometry. Approximate only when the question or interpretation requires it.

Example 10: premature rounding

Suppose \(r=\sqrt{10}\) and we need \(A=\pi r^2\). Exact substitution gives

\[ A=\pi(\sqrt{10})^2=10\pi. \]

Only then do we report \(A\approx31.416\) square units if three decimal places are wanted. Replacing \(r\) first by \(3.16\) gives \(\pi(3.16)^2\approx31.365\), a larger rounding error. Exact intermediate work is both shorter and more accurate.

A check need not repeat the same method. Useful independent checks include:

  • substitute a proposed solution into the original equation;
  • estimate sign and magnitude;
  • compare units;
  • use a second representation, such as a graph;
  • test a limiting or boundary case;
  • expand a factorization or factor an expansion;
  • use exact symbolic computation after the manual derivation.

7. Finding the first invalid step

Later consequences of an error may be correctly derived from a false line. Fixing the last line therefore misses the cause. Locate the first adjacent pair for which the claimed relation fails.

Audit 1: distributing a power

\[ (a+b)^2=a^2+b^2. \]

The first line is already false in general. Multiplication gives \((a+b)^2=a^2+2ab+b^2\). With \(a=b=1\), the false law predicts \(4=2\).

Audit 2: cancelling terms across addition

\[ \frac{x+2}{x}=2. \]

The cancellation is invalid because \(x\) is not a factor of the entire numerator. For \(x\ne0\), the correct form is \(1+2/x\).

Audit 3: dividing by an unknown zero

From \(ab=ac\), a learner writes \(b=c\). The implication is valid only if \(a\ne0\). If \(a=0\), the original equality holds for every \(b,c\). The complete conclusion is \(a=0\) or \(b=c\).

Audit 4: reversing an inequality incorrectly

From \(-2x<6\), the learner writes \(x< -3\). Dividing by a negative number reverses order, so \(x>-3\). A check with \(x=0\) confirms the corrected solution and refutes the proposed one.

Audit 5: taking an even root

From \(x^2=25\), the learner writes \(x=5\). The nonnegative symbol \(\sqrt{25}\) is \(5\), but solving requires both numbers whose square is \(25\): \(x=\pm5\).

Audit 6: treating a radical as linear

The claim \(\sqrt{x^2}=x\) fails at \(x=-1\). The correct real identity is \(\sqrt{x^2}=|x|\).

Audit 7: cross-multiplying through zero denominators

From \(1/(x-1)=2/(x^2-1)\), cross multiplication can be used only after recording \(x\ne1,-1\). Multiplication gives \(x^2-1=2x-2\), hence \((x-1)^2=0\) and candidate \(x=1\)—but that value is excluded. The original equation has no solution.

Audit 8: replacing implication by equivalence

From \(x=2\), squaring yields \(x^2=4\). Writing \(\Leftrightarrow\) adds \(x=-2\) to the reverse direction. The symbol, not the arithmetic, is the first error.

Audit 9: decimal equality

Writing \(1/7=0.143\) is false. The correct statement is \(1/7\approx0.143\) to three decimal places; the absolute error is about \(1.43\times10^{-4}\).

Audit 10: proving by a picture

A graph appears to show that \(x^4+1\) has no real roots. The picture supports a conjecture but cannot inspect every real input. The proof is that \(x^4\ge0\), so \(x^4+1\ge1>0\) for every real \(x\).

Guided audit

A learner solves \(\sqrt{2x+3}=x\) as follows:

\[ \sqrt{2x+3}=x \Leftrightarrow2x+3=x^2 \Leftrightarrow(x-3)(x+1)=0 \Leftrightarrow x=3\text{ or }x=-1. \]

Identify the first incorrect relation and complete the solution.

Ask whether squaring preserves both directions without an added sign condition.

The first \(\Leftrightarrow\) is unjustified. Squaring gives only \(\sqrt{2x+3}=x\Rightarrow2x+3=x^2\) unless \(x\ge0\) is retained. Factoring gives candidates \(3\) and \(-1\). Substitution yields \(\sqrt9=3\) for \(x=3\), while \(\sqrt1=1\ne-1\). The solution set is \(\{3\}\).

8. Common mistakes and how to repair them

Mistake: the equal sign as punctuation

Writing \(18/3=6+4=10\) falsely claims \(18/3=10\). Put separate computations on separate lines, or write \(18/3+4=6+4=10\) if that was the intended expression.

Mistake: checking only the derived equation

If a candidate came from squaring or clearing denominators, satisfying the new equation is not enough. Substitute into the original equation, including its original domain.

Mistake: citing a rule without its conditions

“Cancel \(x\)” is incomplete. State that \(x\ne0\) and identify the common multiplicative factor. Terms separated by addition do not cancel.

Mistake: more symbols means more rigor

A line of unexplained arrows can hide direction errors. Use symbols only when the relation is exact, and use a short sentence when the logical role would otherwise be ambiguous.

9. Guided practice: write the argument, not only the answer

Guided exercise 1: aligned arithmetic

Evaluate \(2(3-7)^2-5\) in an aligned equality chain, changing one feature per line.

\[ \begin{aligned} 2(3-7)^2-5 &=2(-4)^2-5 &&\text{(parentheses)}\\ &=2(16)-5 &&\text{(power)}\\ &=32-5 &&\text{(multiplication)}\\ &=27. &&\text{(subtraction)} \end{aligned} \]

Guided exercise 2: condition for cancellation

Complete a correct statement beginning with \(a(x-2)=a(x+5)\).

Subtracting gives \(-2a=5a\), so \(7a=0\) and therefore \(a=0\). If one instead cancels \(a\), one must assume \(a\ne0\), but that assumption makes the resulting \(-2=5\) impossible. Thus the original equation holds for every \(x\) when \(a=0\) and for no \(x\) when \(a\ne0\).

Guided exercise 3: exact and approximate reporting

The diagonal of a unit square is \(\sqrt2\). Report it exactly, to three decimal places, and with a simple interval that certifies the rounding.

The exact length is \(\sqrt2\). Since \(1.414^2=1.999396<2\) and \(1.415^2=2.002225>2\), we know \(1.414<\sqrt2<1.415\). Hence \(\sqrt2\approx1.414\) to three decimal places.

10. Examination strategy clinic

Problem: when does a familiar transformation become valid?

Determine all real pairs \((a,b)\) for which the implication

\[ a^2=b^2\Rightarrow a=b \]

is valid under each additional condition below. Justify necessity and sufficiency.

  1. \(a,b\ge0\);
  2. \(a+b\ne0\);
  3. \(ab>0\);
  4. \(a\ge b\).

Inventory. The base equation factors as \((a-b)(a+b)=0\), so \(a=b\) or \(a=-b\). Any sufficient condition must eliminate the second branch unless it overlaps the first at zero.

Case 1. If \(a,b\ge0\) and \(a=-b\), then both must be zero, which also gives \(a=b\). The implication is valid.

Case 2. The condition \(a+b\ne0\) directly eliminates the \(a=-b\) factor, so \(a-b=0\). The implication is valid.

Case 3. If \(ab>0\), the numbers have the same nonzero sign. The branch \(a=-b\) would give \(ab=-b^2<0\), impossible. The implication is valid.

Case 4. The condition \(a\ge b\) is not enough: \(a=1\), \(b=-1\) satisfies \(a^2=b^2\) and \(a\ge b\) but not \(a=b\). A counterexample settles the question.

Strategy lesson. Factor first, list all branches, and ask what each extra condition eliminates. This is faster and more reliable than testing random pairs. Necessity questions often require a counterexample; sufficiency questions require an argument covering every allowed case.

11. Exercises by purpose and difficulty

Attempt all problems before opening the answer block. Problems 1–5 emphasize fluency, 6–11 connected reasoning, 12–15 synthesis, and 16–18 honors-level argument.

  1. Classify \(x^2+1\), \(x^2+1=0\), and \(x^2+1>0\) over the reals.

  2. Replace the incorrect symbol in \(\pi=3.14159\).

  3. Solve \(7-2x=19\) using an equivalence chain and check the solution.

  4. State the real domain of \(\sqrt{5-x}/(x+2)\).

  5. Write \(5/6\) to four decimal places using correct notation.

  6. Explain why \(x^2-9=(x-3)(x+3)\) is an identity but \(x^2-9=0\) is not.

  7. Identify the first invalid step: \(2/x=3/x\Leftrightarrow2=3\), so no solution.

  8. Give a counterexample to \((a-b)^2=a^2-b^2\) and state the correct expansion.

  9. Decide which direction is valid between \(x>2\) and \(x^2>4\) over \(\mathbb R\).

  10. Solve \(\sqrt{x+6}=x\) and mark every implication-only step.

  11. A learner writes \((x^2-4)/(x-2)=x+2\). Supply the missing condition and explain its graphical meaning.

  12. Solve \(a(x-1)=2a\) for real \(x\), classifying all real values of \(a\).

  13. Find all real \(x\) satisfying \(\sqrt{x^2}=x-2\).

  14. Critique: “The graphs of two expressions agree for \(-10\le x\le10\), so the expressions are identical.” Give two distinct reasons the conclusion may fail.

  15. Write a valid argument proving that the square of every odd integer is odd.

  16. Determine all real \(c\) for which \(x^2=c\Leftrightarrow x=\sqrt c\) is a true statement for every real \(x\) under consideration. Clarify the universe.

  17. Find the first invalid step in the classic false proof obtained from \(a=b\ne0\): \(a^2=ab\), \(a^2-b^2=ab-b^2\), \((a-b)(a+b)=b(a-b)\), \(a+b=b\), \(2b=b\), \(2=1\). Explain why all earlier lines are valid.

  18. Prove or disprove: if applying the same real-valued function to both sides of an equation preserves equality, then the operation also preserves the solution set.

Concise answers

  1. Expression; conditional equation with no real solution; true predicate (and universally true statement if quantified over all reals).
  2. \(\pi\approx3.14159\).
  3. \(7-2x=19\Leftrightarrow-2x=12\Leftrightarrow x=-6\); substitution gives \(19\).
  4. \(x\le5\) and \(x\ne-2\), or \((-\infty,-2)\cup(-2,5]\).
  5. \(5/6=0.8333\ldots\approx0.8333\) to four decimal places.
  6. Expansion proves the first equality for every real \(x\); the second is true only at \(x=\pm3\).
  7. The first equivalence silently multiplies by \(x\). It needs \(x\ne0\), already forced by the original fractions; with that condition the conclusion of no solution is correct.
  8. \(a=2,b=1\) gives \(1\ne3\); \((a-b)^2=a^2-2ab+b^2\).
  9. \(x>2\Rightarrow x^2>4\) is valid. The reverse fails at \(x=-3\).
  10. The solution is \(x=3\): squaring gives \(x+6=x^2\), candidates \(3,-2\), and substitution rejects \(-2\).
  11. \(x\ne2\); the rational graph is the line \(y=x+2\) with a hole at \((2,4)\).
  12. If \(a=0\), every real \(x\) is a solution. If \(a\ne0\), \(x=3\).
  13. Since \(\sqrt{x^2}=|x|\), solve \(|x|=x-2\). The right side must be nonnegative, so \(x\ge2\) and then \(|x|=x\), giving \(x=x-2\), impossible. No solution.
  14. A finite window misses behavior outside it, and finite-resolution plots can hide small differences, holes, or isolated exceptions.
  15. See the full solution below.
  16. With universe \(x\ge0\), the equivalence is true for every \(c\ge0\) when read as a solution statement; over all real \(x\), no positive \(c\) works because \(x=-\sqrt c\) is lost, while \(c=0\) does work.
  17. Cancelling \(a-b\) is invalid because \(a=b\), so \(a-b=0\). Earlier lines are equality-preserving algebra.
  18. False. Equality is preserved by every function: \(u=v\Rightarrow f(u)=f(v)\). Solution sets are preserved only when the function is injective on the relevant values. Squaring loses injectivity: \(x=-1\) implies \(x^2=1\), whose solution set also contains \(1\).

Selected full solutions

The original equation requires \(x+6\ge0\) and also \(x\ge0\). Squaring is therefore recorded as implication:

\[ \sqrt{x+6}=x\Rightarrow x+6=x^2 \Leftrightarrow x^2-x-6=0 \Leftrightarrow(x-3)(x+2)=0. \]

The candidates are \(3\) and \(-2\). The sign condition already excludes \(-2\), and substitution confirms \(\sqrt9=3\). Thus the solution set is \(\{3\}\).

Let \(n\) be an arbitrary odd integer. By definition, \(n=2k+1\) for some integer \(k\). Then

\[ n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1. \]

Because integers are closed under multiplication and addition, \(2k^2+2k\) is an integer. Hence \(n^2\) has the form \(2m+1\) for an integer \(m\) and is odd. The word “arbitrary” matters: the proof did not depend on a special odd integer.

Fix \(c\ge0\). Over the universe \(x\in\mathbb R\), the equation \(x^2=c\) has solutions \(x=\pm\sqrt c\). The statement \(x=\sqrt c\) retains only the nonnegative one. They are equivalent only when the two roots coincide, which occurs at \(c=0\). If instead the universe is restricted in advance to \(x\ge0\), then the principal-square-root definition gives \(x^2=c\Leftrightarrow x=\sqrt c\) for every \(c\ge0\). For \(c<0\), \(\sqrt c\) is not a real number, so the displayed real statement is not an admissible equivalence.

12. Cumulative retrieval

This first chapter has no earlier book chapter to retrieve, so use prerequisite ideas in a new logical form.

Fractions plus domain reasoning

Solve

\[ \frac{1}{x} + \frac{1}{x+1}=\frac32 \]

over the reals. Record restrictions before clearing denominators and check every candidate.

The restrictions are \(x\ne0,-1\). Multiplying by \(2x(x+1)\) is reversible on that domain:

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

so \(3x^2-x-2=0=(3x+2)(x-1)\). The candidates are \(x=-2/3\) and \(x=1\), neither excluded. Substitution gives \(-3/2+3=3/2\) and \(1+1/2=3/2\), respectively. The solution set is \(\{-2/3,1\}\).

13. Topic checkpoint

The checkpoint samples classification, equivalence, domain, exactness, and argument repair. Treat explanations as part of the answer.

14. Summary and next step

  • Expressions name objects; equations, inequalities, and quantified predicates make claims.
  • Equality, approximation, implication, and equivalence have different logical jobs.
  • Domain restrictions and parameter cases must be written before an operation hides them.
  • A derivation changes one thing per line and identifies nontrivial reasons.
  • Squaring, applying non-injective functions, and clearing expressions under uncertain conditions may produce only implications.
  • Examples support intuition; proofs establish general claims; one valid counterexample refutes a universal claim.
  • Audit the first invalid step and verify the final answer in the original problem.

Next, use these habits in the readiness diagnostic and repair chapter. The diagnostic measures separate subskills so that practice is targeted rather than repetitive.

References and further study

  • Richard Hammack, Book of Proof, especially the opening chapters on statements and proof language.
  • OpenStax, Algebra and Trigonometry 2e, for prerequisite algebra and solution checking.
  • Daniel Velleman, How to Prove It, for a deeper treatment of logical form and proof construction.
  • George Pólya, How to Solve It, for planning, checking, and reflection habits.
  • SymPy documentation, Solvers, for interpreting symbolic solution workflows.

Using this lesson with fcmath and SymPy

Computation belongs after the manual argument. The first example checks an identity; the second distinguishes a candidate-producing transformation from a verified solution set.

import sympy as sp

x = sp.symbols("x", real=True)
identity_difference = sp.expand((x + 1) ** 2 - (x**2 + 2 * x + 1))
identity_difference

\(\displaystyle 0\)

An output of 0 shows that the difference simplifies to zero symbolically. It supports the algebraic identity; the distributive-law derivation explains why it is true.

original = sp.sqrt(x + 2) - x
candidates = sp.solve(sp.Eq(x + 2, x**2), x)
verified = [value for value in candidates if sp.simplify(original.subs(x, value)) == 0]
candidates, verified
([-1, 2], [2])

The squared equation produces [-1, 2], while substitution in the original expression retains only [2]. This is the computational version of the implication audit, not a replacement for it.

A reusable future fcmath transformation-audit helper should record the original domain, operation, direction, new candidates, and verification result. Until that API is independently reviewed, keep those fields explicitly in your written work.

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