Chapter 5D: Mixed fractions, cyclic reasoning and error audits

Bracketed sums are whole objects

A large division sign between two bracketed sums divides one complete result by another. For (1/2+1/3)÷(1/2−1/3), the sums are 5/6 and 1/6, so the quotient is 5. In algebra, combine each bracket first or use a common denominator deliberately; never flip only the final term of a sum.

Multiplication and division share priority

At one bracket level, process multiplication and division from left to right unless parentheses say otherwise. A÷B×C means (A÷B)×C, not A÷(B×C). For 12÷3×2 the result is 8, while 12÷(3×2)=2. A fraction bar groups its complete numerator and denominator; draw the bars and brackets before transforming a mixed expression.

Use a shared denominator inside a quotient

Let U=x/(x+y)+y/(x−y) and V=x/(x−y)−y/(x+y). Both simplify to (x²+y²)/(x²−y²). Thus U/V=1 where x²−y²≠0 and V≠0. For real x,y the first condition already prevents x²+y²=0, but state the divisor condition before using that fact. Equal-looking final brackets do not justify ignoring their original domains.

A quadratic/quartic ladder prevents excessive expansion

(1−x+x²)(1+x+x²)=1+x²+x⁴: write it as (1+x²)²−x². Therefore 1/(1−x+x²)−1/(1+x+x²)=2x/(1+x²+x⁴). The denominator is positive for real x because (x²+1/2)²+3/4>0. The related successive difference 1/(x−1)−1/(x+1)−2/(x²+1)−4/(x⁴+1)=8/(x⁸−1), for real x≠±1. Changing the last minus to plus changes the numerator to 8x⁴.

Read the exact cyclic denominator pattern

For D=(x−y)(y−z)(z−x), the sum y/[(x−y)(y−z)] + x/[(z−x)(x−y)] + z/[(y−z)(z−x)] has numerator y(z−x)+x(y−z)+z(x−y)=0. Changing the numerators to x−y,y−z,z−x over the complementary two-factor denominators instead gives a sum of three squares over D, not generally zero. Pairwise differences must all be nonzero.

Recognise cube expressions before dividing

Use x³+y³+3xy(x+y)=(x+y)³ and x³−y³−3xy(x−y)=(x−y)³. Also (x+y)²−4xy=(x−y)² and (x−y)²+4xy=(x+y)². Replacing these complete expressions turns a complicated quotient into x²−y², for x−y≠0 and x+y≠0. A companion using a,b in an analogous expression must keep those letters consistent; a variable name is not decoration.

A multi-part task reuses data but each part needs its own reasoning

For F=(a⁴−b⁴)/(a−b)², factor to (a+b)(a²+b²)/(a−b), a≠b. Multiplying F by (a−b)/(a³+b³) and (a+b)/(a³+b³) yields (a²+b²)/(a²−ab+b²)² on the common original domain. For a different part, divide F by [(a³+a²b+ab²+b³)/((a+b)²−4ab)] and obtain a−b; adding a²/(a+b) gives (2a²−b²)/(a+b), with all original denominators and the divisor nonzero.

A requested proof can itself be false

Printed creative 6(b) defines A=1/(1−x+x²), B=1/(1+x+x²), C=1/(1+x²+x⁴), then requests 1/A−1/B−2x/C=0. At x=1, A=1 and B=C=1/3, so the stated left side is −8, not zero. The related identity A−B−2xC=0 is valid and follows from the quadratic/quartic factorisation. Keep the printed claim and correction distinct. Printed sample 3(iii) is a different case: its incorrect quotient is an assertion to reject inside a multiple-choice task, not a formula to accept.

Worked example

For A=1/(1−x+x²), B=1/(1+x+x²), C=1/(1+x²+x⁴), factor the common denominator to show A−B=2xC. At x=1 this reads 1−1/3=2/3. The printed reciprocal claim instead gives 1−3−6=−8, so it cannot be an identity.

1. Record every original nonzero denominator and, for division, the nonzero divisor. Identify the operation and factor complete expressions.
2. Show a common denominator for addition/subtraction or a reciprocal for division. Cancel only complete common factors and preserve brackets.
3. Check a permitted numerical input and retain the original exclusions. Compare full written reasoning with the model; the automatic checker handles only the requested finite answer.

Real and complex domains differ

These lessons use real variables unless stated otherwise. Positivity proofs for real quadratics do not claim they have no complex roots.

An assertion to reject is different from an erroneous solution

The sample quotient assertion is deliberately false inside a question. The creative reciprocal claim and the changed example numerator are source conflicts, recorded and explained rather than silently fixed.

Explore the mathematical figures

Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.

Four checks keep a mixed fraction valid

A · Original domain

Include original denominators and any zero divisor.

B · Grouping and order

A fraction bar groups; multiplication/division share priority.

C · Factors

Show identities and missing multipliers before cancelling.

D · Check

Use permitted values and review the symbolic reasoning.

Original decision aid, not an automatic symbolic grading process or a guarantee that an arbitrary expression is valid.

Explore the mathematical figures

Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.

A cyclic sum needs its exact factor pattern

A · First transformed term

(x−y)(x+y)=x²−y².

B · Second

(y−z)(y+z)=y²−z².

C · Third

(z−x)(z+x)=z²−x²; all square terms cancel.

Original symbolic cycle. Arrow directions show cancellation relationships, not forces or a scaled triangle. Require all three pair sums nonzero.