Simplest Form
A fraction is in simplest form when the numerator and denominator share no common factor other than 1.
Formula
\frac{a}{b} \text{ where } \gcd(a,b) = 1
Definition
A fraction is in simplest form (also called lowest terms) when you cannot divide both the top and bottom by any number other than $1$, the smallest, cleanest version of the fraction. Formally, a fraction $a/b$ is in simplest form when $\gcd(a, b) = 1$, meaning the numerator and denominator are coprime, and reducing means dividing both by their GCD: $(a/\gcd)/(b/\gcd)$. The simplest form of a rational number $a/b$ is the unique representative $(a', b')$ of its equivalence class with $b' > 0$ and $\gcd(a', b') = 1$; existence and uniqueness follow from the fundamental theorem of arithmetic and properties of the GCD.
Example
$6/8$ is NOT in simplest form because you can divide both by $2$ to get $3/4$, which IS simplest since $3$ and $4$ share no common factor other than $1$. To reduce $36/48$: since $36 = 2^2 \times 3^2$ and $48 = 2^4 \times 3$, $\gcd(36,48) = 2^2 \times 3 = 12$, so $36/48 = (36/12)/(48/12) = 3/4$. The idea generalizes beyond the integers: for a Gaussian integer fraction $(a+bi)/(c+di)$ in $\mathbb{Z}[i]$, the analogue of simplest form requires dividing by the GCD in $\mathbb{Z}[i]$ (a Euclidean domain), using the norm function $N(a+bi) = a^2+b^2$ as the "size" in the Euclidean algorithm.
Key Insight
Simplest form makes fractions easier to understand and compare, it's like giving someone your address as "123 Main St" instead of "246/2 Main St," same place, much clearer. The Euclidean algorithm computes the GCD efficiently even for large numbers ($\gcd(48, 36) = \gcd(36, 12) = \gcd(12, 0) = 12$), beating the approach of listing all factors, and it is the basis for modern cryptographic key generation. Coprimality ($\gcd = 1$) also directly controls the period of the decimal expansion: $1/b$ terminates iff $b$ has only factors of $2$ and $5$, otherwise the period length equals the multiplicative order of $10$ modulo ($b$ after removing $2$s and $5$s).