Algebra Proof Collection

Five Elegant Algebra Proofs

The "why" behind the rules you memorized
Proof 1 Why Negative × Negative = Positive

Most people learn "$(-1)(-1) = 1$" as a rule to memorize. But it can be proven using only the distributive property and the fact that anything times zero is zero. No circular reasoning needed.

We want to prove
$$(-a)(-b) = ab \qquad \text{for all real numbers } a, b$$

We'll prove the core case $(-1)(-1) = 1$ first. The general case follows immediately since $(-a)(-b) = (-1)(a)(-1)(b) = (-1)(-1)(ab)$.

Step 1 — Start with a known fact

We know that any number plus its additive inverse equals zero:

$$1 + (-1) = 0$$
Step 2 — Multiply both sides by $(-1)$

Multiplying both sides of an equation by the same number preserves equality:

$$(-1)\bigl[1 + (-1)\bigr] = (-1)(0)$$
Step 3 — Simplify the right side

Anything times zero is zero:

$$(-1)\bigl[1 + (-1)\bigr] = 0$$
Step 4 — Distribute on the left side

Apply the distributive property $a(b+c) = ab + ac$:

$$(-1)(1) + (-1)(-1) = 0$$
Step 5 — Simplify and solve

We know $(-1)(1) = -1$. Substituting:

$$-1 + (-1)(-1) = 0$$

The only number that makes this true is $(-1)(-1) = 1$.

The general case

For any $a, b > 0$: $(-a)(-b) = (-1)(a)(-1)(b) = (-1)(-1) \cdot ab = 1 \cdot ab = ab$. The proof rests on just three axioms: the distributive property, the existence of additive inverses, and $a \cdot 0 = 0$.

■ Q.E.D.

Proof 2 The Sum & Difference of Cubes

These two factoring formulas seem to appear from nowhere. But they can be verified by simply multiplying the right side back out and watching all the middle terms cancel.

The formulas
$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$

Proof of the difference of cubes (the sum follows the same pattern):

Step 1 — Expand $(a - b)(a^2 + ab + b^2)$

Distribute $a$ across the trinomial, then distribute $(-b)$:

$$a \cdot a^2 + a \cdot ab + a \cdot b^2 + (-b) \cdot a^2 + (-b) \cdot ab + (-b) \cdot b^2$$
Step 2 — Simplify each term
$$a^3 + a^2 b + ab^2 - a^2 b - ab^2 - b^3$$
Step 3 — Cancel the middle terms

Watch the magic: $+a^2b$ cancels with $-a^2b$, and $+ab^2$ cancels with $-ab^2$:

$$\cancel{a^2 b} - \cancel{a^2 b} + \cancel{ab^2} - \cancel{ab^2} = 0$$

All that survives is:

$$a^3 - b^3 \qquad \checkmark$$
Why the middle terms always cancel

The trinomial $a^2 + ab + b^2$ is engineered so that when you multiply by $(a - b)$, the cross terms appear in equal-and-opposite pairs. The same trick works for higher powers: $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$ for any positive integer $n$.

■ Q.E.D.

Proof 3 The Distance Formula & Equation of a Circle

The distance formula isn't a new idea — it's the Pythagorean Theorem in disguise. And the equation of a circle follows immediately from it.

Part A — The Distance Formula
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Step 1 — Set up a right triangle

Given two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, draw a horizontal line from $P_1$ and a vertical line from $P_2$. They meet at the point $(x_2, y_1)$, forming a right triangle. The horizontal leg has length $|x_2 - x_1|$ and the vertical leg has length $|y_2 - y_1|$.

Step 2 — Apply the Pythagorean Theorem

The distance $d$ between $P_1$ and $P_2$ is the hypotenuse:

$$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Taking the positive square root:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \qquad \checkmark$$

Part B — The Equation of a Circle
$$(x - h)^2 + (y - k)^2 = r^2$$
Step 3 — Define a circle

A circle is the set of all points $(x, y)$ that are exactly distance $r$ from a fixed center $(h, k)$. In other words, for every point on the circle:

$$\text{distance from } (x,y) \text{ to } (h,k) = r$$
Step 4 — Substitute the distance formula

Replace "distance" with the formula from Part A:

$$\sqrt{(x - h)^2 + (y - k)^2} = r$$
Step 5 — Square both sides
$$(x - h)^2 + (y - k)^2 = r^2 \qquad \checkmark$$
Connecting the dots

The equation of a circle isn't a separate formula to memorize — it literally is the distance formula with $d$ set equal to $r$. If you know the Pythagorean Theorem, you know the equation of a circle.

■ Q.E.D.

Proof 4 Why $\sqrt{2}$ Is Irrational

This is one of the oldest and most famous proofs in mathematics, dating back to ancient Greece. It uses proof by contradiction: assume the opposite of what you want to prove, then show that assumption leads to a logical impossibility.

Theorem
$$\sqrt{2} \text{ cannot be expressed as a fraction } \frac{p}{q} \text{ where } p, q \text{ are integers.}$$
Step 1 — Assume the opposite

Suppose, for the sake of contradiction, that $\sqrt{2}$ is rational. Then we can write:

$$\sqrt{2} = \frac{p}{q}$$

where $p$ and $q$ are integers with no common factors (the fraction is fully reduced).

Step 2 — Square both sides
$$2 = \frac{p^2}{q^2} \qquad \Longrightarrow \qquad p^2 = 2q^2$$
Step 3 — Conclude $p$ is even

Since $p^2 = 2q^2$, we see that $p^2$ is even (it's $2$ times something). But if $p^2$ is even, then $p$ itself must be even. (If $p$ were odd, $p^2$ would also be odd.)

Since $p$ is even, we can write $p = 2m$ for some integer $m$.

Step 4 — Substitute and simplify

Replace $p$ with $2m$ in the equation $p^2 = 2q^2$:

$$(2m)^2 = 2q^2 \qquad \Longrightarrow \qquad 4m^2 = 2q^2 \qquad \Longrightarrow \qquad q^2 = 2m^2$$
Step 5 — Conclude $q$ is also even

By the same reasoning as Step 3, since $q^2 = 2m^2$, we know $q^2$ is even, so $q$ must be even too.

Step 6 — Contradiction!

We've shown both $p$ and $q$ are even. But in Step 1, we said $p/q$ was fully reduced — meaning $p$ and $q$ share no common factors. If both are even, they share the factor $2$. Contradiction.

Our original assumption must be false. Therefore, $\sqrt{2}$ is irrational.

Why this matters

This proof shook ancient Greek mathematics. The Pythagoreans believed all quantities could be expressed as ratios of whole numbers. The discovery that the diagonal of a unit square ($\sqrt{2}$) cannot be so expressed was a genuine philosophical crisis. The same technique proves $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{p}$ for any prime $p$ are all irrational.

■ Q.E.D.

Proof 5 Logarithm Properties from Exponent Rules

The three log laws — product, quotient, and power — aren't separate rules. They are the exponent rules, viewed through the lens of the logarithm. Each proof is just a few lines.

The three log laws
$$\log_b(MN) = \log_b M + \log_b N$$
$$\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N$$
$$\log_b(M^k) = k \cdot \log_b M$$

Let $\log_b M = m$ and $\log_b N = n$. This means $b^m = M$ and $b^n = N$.

Product Rule: $\log_b(MN) = \log_b M + \log_b N$

Proof

Multiply $M$ and $N$ in exponential form:

$$MN = b^m \cdot b^n = b^{m+n}$$

Now take $\log_b$ of both sides:

$$\log_b(MN) = m + n = \log_b M + \log_b N \qquad \checkmark$$

Quotient Rule: $\log_b(M/N) = \log_b M - \log_b N$

Proof

Divide $M$ by $N$ in exponential form:

$$\frac{M}{N} = \frac{b^m}{b^n} = b^{m-n}$$

Take $\log_b$:

$$\log_b\!\left(\frac{M}{N}\right) = m - n = \log_b M - \log_b N \qquad \checkmark$$

Power Rule: $\log_b(M^k) = k \cdot \log_b M$

Proof

Raise $M$ to the power $k$ in exponential form:

$$M^k = (b^m)^k = b^{mk}$$

Take $\log_b$:

$$\log_b(M^k) = mk = k \cdot \log_b M \qquad \checkmark$$
The pattern

Every log law comes from the same three-step template: (1) rewrite using $b^m = M$, (2) apply the corresponding exponent rule, (3) convert back to logarithmic form. Multiplication becomes addition, division becomes subtraction, and powers become multiplication — logarithms are a bridge between the world of multiplication and the world of addition.

■ Q.E.D.