From Arithmetic to Algebra: Free Response

Straight through from the left the product is awkward, so rearrange before multiplying anything.

The commutative law of multiplication lets two factors trade places, which brings the $25$ alongside the $4$:

$4 \times 17 \times 25 = 4 \times 25 \times 17.$

The associative law lets the first two be grouped and multiplied first:

$(4 \times 25) \times 17 = 100 \times 17 = 1700.$

Neither law was invoked because these particular numbers happen to be friendly. Both are identities, holding for every three numbers, so the route is always available; here it happens to produce a factor of $100$ along the way.

A claim about any three numbers is destroyed by a single triple that breaks it, so choose numbers that make both bracketings easy to evaluate. Take $a = 36$, $b = 6$ and $c = 3$.

Bracketing the first pair:

$(36 \div 6) \div 3 = 6 \div 3 = 2.$

Bracketing the second pair instead:

$36 \div (6 \div 3) = 36 \div 2 = 18.$

The two results are $2$ and $18$, so where the brackets sit does change the answer. Division has no associative law, which is exactly why a written chain of divisions is read strictly from the left, while a chain of multiplications may be regrouped at will.

The two directions are not symmetric, and that asymmetry is what makes algebra necessary rather than merely convenient.

To refute, one instance is enough. Part B's claim covered every triple, so a single triple on which the two bracketings differ contradicts it outright, and no further checking adds anything to the verdict.

To confirm, one instance is not enough, and neither is any finite list of them. There are infinitely many triples, so whatever you test leaves infinitely many untested, and agreement across your list could be a coincidence that ends at the very next triple. Testing raises confidence and never reaches certainty.

What does reach certainty is a derivation. The associative law of multiplication,

$(ab)c = a(bc),$

is one of the statements taken as holding for every number, so a rearrangement justified by it is justified for every number at once, with nothing left over to check. That is the whole reason the laws are written with letters instead of collected as a list of instances.

Substitute in parentheses each time, and multiply before adding, exactly as the order of operations requires.

At $t = 1$:

$3(1) + 8 = 11, \qquad 8(1) + 3 = 11.$

At $t = 4$:

$3(4) + 8 = 20, \qquad 8(4) + 3 = 35.$

So the two expressions agree at one of these inputs and disagree at the other.

Each law says exactly one thing, so apply it exactly.

The commutative law of addition, $a + b = b + a$, swaps the two quantities being ADDED. Here those are the term $3t$ and the term $8$:

$3t + 8 = 8 + 3t.$

The commutative law of multiplication, $ab = ba$, swaps the two quantities being MULTIPLIED. Inside the term $3t$ those are $3$ and $t$:

$3t = t \cdot 3.$

Putting that back into the expression gives $t \cdot 3 + 8$.

Notice what neither law did. The $8$ never became a coefficient, and the $3$ never left the letter it multiplies. Both laws move things around within one operation, and neither transfers a number out of one term and into another.

Two expressions are equivalent when they take the same value at EVERY input, so a single input at which they differ settles the question in the negative. Part A supplies one:

$3(4) + 8 = 20, \qquad 8(4) + 3 = 35.$

They differ there, so the two are not equivalent, and the agreement at $t = 1$ does not rescue them. One input where two expressions agree is one input, not a promise about the rest.

Now the law. The commutative law of addition permits the two terms of a sum to trade places, and the commutative law of multiplication permits the two factors of a product to trade places. Both moves stay inside a single operation. Turning $3t + 8$ into $8t + 3$ does neither: it takes the $8$ out of the constant term and makes it a coefficient, and it takes the $3$ off the letter and leaves it standing alone. No law in the lesson performs that move, and part A shows why no law could.

Read each route straight into symbols.

Discount first: one lesson inside a pack costs $c - 5$ dollars, and a pack holds eight of them, so the pack costs

$8(c - 5).$

Discount last: eight lessons at the single-lesson rate cost $8c$ dollars, and the school gives $5$ dollars off each of the eight, so the amount to remove is $8 \times 5 = 40$ dollars:

$8c - 40.$

The distributive law is precisely the statement that these two routes agree, for every value of $c$:

$8(c - 5) = 8c - 8 \times 5 = 8c - 40.$

Route one, discount first. A lesson inside a pack costs $23 - 5 = 18$ dollars, and eight of them cost

$8 \times 18 = 144.$

Route two, discount last. Eight lessons at the single-lesson rate cost $8 \times 23 = 184$ dollars, and the discount removed is $8 \times 5 = 40$ dollars, so the pack costs

$184 - 40 = 144.$

Both routes give a pack price of $144$ dollars. They were always going to agree: the two routes are the two sides of the distributive law, and this shared value at $c = 23$ is one instance of an agreement that holds at every price.

Test it at one value of $c$ against a route already known to price a pack correctly. Take $c = 23$ dollars again:

$8(23) - 5 = 184 - 5 = 179.$

The correct pack price at that rate is $144$ dollars, so the two disagree and the expression is not equivalent to either route in part A. One value is enough, because equivalence would have demanded agreement at every value.

The gap is not random. The two prices differ by

$179 - 144 = 35,$

which is seven lots of $5$ dollars. The discount has been handed back once, for the pack as a whole, when the school promised it on each of the eight lessons. Seven of the eight discounts have gone missing, so a customer charged this price pays $35$ dollars more than the advertised deal.

Choose numbers with nothing special about them, in particular a first factor that is not $1$ and a last number that is not $0$. Take $a = 3$, $b = 2$ and $c = 5$.

The left side resolves the bracket first, then multiplies:

$3(2 + 5) = 3 \times 7 = 21.$

The right side multiplies only the first term inside, then adds:

$3 \times 2 + 5 = 6 + 5 = 11.$

The values $21$ and $11$ are different, so this single triple refutes a rule that claimed to hold for all numbers. Nothing further needs testing.

The correct statement is the distributive law:

$a(b + c) = ab + ac.$

The factor outside the bracket reaches BOTH terms inside it. The student's version,

$a(b + c) = ab + c,$

carries that factor into the first term and drops it at the second. Setting the two right sides beside each other, they differ by exactly the multiplication the $c$ never received:

$(ab + ac) - (ab + c) = ac - c.$

So this is not a near miss with a small slip in it. A whole product is absent, and the amount missing grows as $a$ and $c$ grow.

The two checks were not unlucky. They were checks that could not have gone the other way, and it is worth pinning down exactly which triples have that property.

Compare the two right-hand sides, $ab + ac$ and $ab + c$. Both contain $ab$, so subtract it from each. The two sides are equal exactly when

$ac = c.$

That works in both directions: if $ac = c$, then adding $ab$ to each gives two equal sides; and if the two sides are equal, then subtracting $ab$ from each leaves $ac = c$. So this one condition decides every triple.

Now look at what the student chose. The first check used $a = 1$, and multiplying by $1$ returns the same number:

$1 \times 8 = 8.$

The second used $c = 0$, and multiplying by $0$ gives $0$ either way:

$7 \times 0 = 0.$

Each triple met the condition, so each was guaranteed to come out equal before any arithmetic began. A test that cannot fail carries no information: it would have agreed even if the rule had been far more wrong than it is. The tests worth running are the ones that could have gone the other way, which is why part A avoided both of these features.

The claim is about every number, so no amount of checking can establish it. The argument has to run in letters.

Start from a fact about zero alone, $0 + 0 = 0$, and multiply both sides by $a$. The distributive law opens the bracket:

$a \times 0 = a(0 + 0) = a \times 0 + a \times 0.$

Write $u$ for the quantity $a \times 0$. What that line says is

$u = u + u.$

Now add $-u$, the additive inverse of $u$, to both sides. On the left, the inverse law gives $u + (-u) = 0$. On the right, regroup first, which the associative law permits:

$(u + u) + (-u) = u + \bigl(u + (-u)\bigr).$

The inner bracket is $0$ by the inverse law, and $u + 0 = u$ by the identity law, so the right side is just $u$. The two sides were equal before the addition, so they are equal after it:

$0 = u = a \times 0.$

No step named a particular number in place of $a$, so this holds for every $a$ at once.

Suppose, to see what goes wrong, that zero did have a multiplicative inverse. That would mean some number $k$ satisfies

$0 \times k = 1,$

since a multiplicative inverse is by definition a partner that multiplies with it to give $1$.

But part A settles that same product. Commuting the two factors first, $0 \times k = k \times 0$, and part A applies to $k$ as much as to any other number:

$k \times 0 = 0.$

The one product would then have to be both $1$ and $0$, which would put $1 = 0$. That is false, so the supposition is impossible: no number is a multiplicative inverse of zero.

The restriction printed beside the law is therefore not tidiness or taste. It is the only thing the other laws leave room for, and a system that dropped it would contradict itself.

The two statements have the same reach and a different standing, and telling them apart is what working from first principles means in practice.

$a + 0 = a$ is a law. Nothing in the lesson derives it; it is one of the statements the subject starts from, written down because it records how addition and zero actually behave. Its job is to be assumed.

The statement in part A is a theorem. Nothing about multiplying by zero was assumed anywhere. The conclusion came out of the distributive, associative, inverse and identity laws working together, so its truth is inherited from theirs.

That difference decides what an attack on each would have to look like. To unseat the law you would have to work in a different number system, one whose starting assumptions are not these, and the ordinary numbers would no longer be what you were talking about. To unseat the theorem while keeping the laws you would have to find a broken step in the chain, because so long as the laws hold, a line such as

$a(0 + 0) = a \times 0 + a \times 0$

is not a separate belief that could turn out false on its own. This is why a first-principles subject keeps its assumptions short: everything else is on the hook for a proof, and nothing that has been proved can fail independently of what it was proved from.