Pre-Calc Diagnostic: State True or False (Stewart Problem 10)

Direct answer. The verdicts: (a) (p + q)² = p² + q² is FALSE (the missing middle term is 2pq); (b) ab = a b is TRUE (radical product rule, nonneg a, b); (c) a² + b² = a + b is FALSE (try the 3-4-5 triangle: √25 = 5, not 7); (d) 1 + TCC = 1 + T is FALSE (correct: 1C + T); (e) 1x − y = 1x − 1y is FALSE (cannot split across subtraction); (f) 1/xa/x − b/x = 1a − b is TRUE (the x's legitimately cancel). Five out of six are false — by design. This problem catches the misconceptions students smuggle into Calc.

The big-picture lesson before any individual answer

Every false identity on this problem has the same structure: a student tried to distribute an operation across an operation it doesn't distribute over. Squaring doesn't distribute over addition. The square root doesn't distribute over addition or subtraction. A fraction can be split when there's addition in the numerator, but not when there's addition or subtraction in the denominator. Once you internalize that one principle, every part of Stewart's Problem 10 takes about five seconds. The remaining work is just supplying the counter-example.

The two TRUE statements are also instructive. They are true because the operation involved does respect the structure: multiplication distributes inside a square root over multiplication, and cancellation works whenever the same nonzero factor appears top and bottom. Both of these students often distrust — they look "too easy" — but they are valid.

(a) (p + q)² = p² + q²   FALSE

Squaring a binomial means multiplying the binomial by itself. The cross-terms do not vanish:

(p + q)² = (p + q)(p + q) = p² + 2pq + q²

The missing term is 2pq. Counter-example with the smallest possible numbers: let p = 1 and q = 1. Then (1 + 1)² = 4 but 1² + 1² = 2. The two sides are equal only when 2pq = 0, i.e. when at least one of p or q is zero. This is the single most-tested algebra mistake in all of pre-calculus. The FOIL drill exists for one reason: to prevent this exact error. Drill it until it's automatic.

(b) ab = a b   TRUE (for a, b ≥ 0)

The radical product rule. It follows immediately from the definition of a square root: if you square both sides, you get ab = ab in both cases. As long as a and b are nonnegative, you can split the radical across a product or combine two radicals into one. We use this constantly when simplifying — e.g., 50 = 25 · 2 = 52.

The one trap: the rule does not survive on negatives. (−1)(−1) = 1 = 1, but −1 · −1 is not defined over the real numbers. Stewart's diagnostic restricts to a, b ≥ 0, so the identity holds as stated.

(c) a² + b² = a + b   FALSE

The square root does not distribute over addition. There is no rule for "splitting" stuff + stuff into two separate radicals. The cleanest counter-example is the 3-4-5 right triangle:

3² + 4² = 9 + 16 = 25 = 5

a + b = 3 + 4 = 7

So 5 ≠ 7, and the equation is false. Notice the deeper geometric meaning: this is exactly why the Pythagorean theorem exists. The hypotenuse a² + b² is never the sum of the legs unless one leg is zero. The shortest distance between two points in the plane is a straight line, not a right-angle path, and that fact is structurally identical to a² + b² ≤ a + b.

(d) 1 + TCC = 1 + T   FALSE

A fraction with a sum in the numerator splits over the denominator — but you must split every term, not just the one that looks easy:

1 + TCC = 1C + TCC = 1C + T

The correct simplification is 1C + T, not 1 + T. The student who wrote "1 + T" dropped the 1/C because the C's "obviously" canceled — but they only cancel in the term that contains a C. The lone 1 in the numerator does not contain a C, so there is nothing to cancel.

The verbal rule worth memorizing: you can split a fraction over addition in the numerator if you split every term. You cannot do anything similar with addition in the denominator.

(e) 1x − y = 1x − 1y   FALSE

Same family as part (c): you cannot split across subtraction in the denominator. The denominator is one expression; it cannot be torn apart. To combine the right side over a common denominator:

1x − 1y = y − xxy

which is not 1/(x − y). Quick numerical check: with x = 2, y = 1, the left side is 1/(2 − 1) = 1, the right side is 1/2 − 1 = −0.5. They aren't even close.

The mnemonic: fractions split across addition in the numerator; never across addition or subtraction in the denominator.

(f) 1/x(a − b)/x = 1a − b   TRUE

The complex fraction. Two ways to verify it works:

Method 1 — invert and multiply. Dividing by a fraction means multiplying by its reciprocal:

1/x(a − b)/x = 1x · xa − b = xx(a − b) = 1a − b

Method 2 — clear the inner denominators. Multiply numerator and denominator of the big fraction by x:

1/x · x((a − b)/x) · x = 1a − b

Both routes give the same answer. The x's legitimately cancel because they appear as a multiplicative factor in both numerator and denominator — nothing is being torn across an addition or subtraction. Many students reflexively distrust this kind of cancellation after being burned by part (d) and part (e), but cancellation across a clean multiplication is exactly the case where it works.

The four mistake-shapes worth memorizing forever

• Don't square term-by-term. (p + q)² ≠ p² + q². Use FOIL or the binomial formula.
• Don't take a square root term-by-term. a² + b² ≠ a + b. Counter-example: 3, 4, 5.
• Split a sum over a denominator? Every term gets split. (1 + TC)/C = 1/C + T, not 1 + T.
• Never split a fraction across addition or subtraction in the DENOMINATOR. 1/(x − y) ≠ 1/x − 1/y.

If a pre-calc student walks into AP Calc with those four shapes hard-wired, they will recover hours of points per exam. Calc is hard because it asks you to manipulate complicated expressions quickly — a student who has to think about whether (p + q)² = p² + 2pq + q² will never finish a free-response section in the allotted time.

How SOMATH teaches the misconceptions diagnostic

We open every new pre-calc student's first session with a six-line drill very close to Stewart's. They write T or F next to each one, and — critically — they have to supply the counter-example or the underlying property out loud before they explain. The drill takes five minutes. It tells us exactly which of the four mistake-shapes are still live for that student, and we spend the rest of the unit fixing the live ones. From our classroom on 226 W 79th St we run small groups of 3–5, every lesson on whiteboards, every counter-example written out. Every new student gets a free 30-minute evaluation and a written diagnostic within 48 hours, even if they don't enroll.

FAQ

Why is (p + q)² = p² + q² false?

The middle term 2pq is missing. The correct expansion is (p + q)² = p² + 2pq + q².

Is ab = a b always true?

Yes, for nonnegative a and b. The rule fails on negatives because the radical is undefined over the reals.

Why is a² + b² = a + b false?

Square root does not distribute over addition. Counter-example: a = 3, b = 4 gives 5, not 7. This is the Pythagorean theorem in disguise.

Can I split (1 + TC)/C as 1 + T?

No. Split every term: 1/C + T. Dropping the 1/C is the mistake.

Why is 1/(x − y) ≠ 1/x − 1/y?

You cannot split a fraction across subtraction in the denominator. The denominator is one unit. Quick check: x = 2, y = 1 gives 1 vs. −0.5.

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