11.1 Sets (Kümeler), Cartesian Products, Binary Operations (İşlem) & Modular Arithmetic (Modüler Aritmetik)

Sets (Kümeler), Cartesian Products, Binary Operations & Modular Arithmetic

On the TR-YÖS, set theory, abstract binary operations (işlem), and modular arithmetic (modüler aritmetik) bridge foundational arithmetic and structural algebra. Questions in this domain rarely ask for long proofs; instead, they test whether you can immediately translate element constraints into powers of $2$, apply inclusion-exclusion without double-counting, extract identity and inverse elements from a custom algebraic rule, and collapse giant exponents using modular periods.

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1. Sets (Kümeler) & Constrained Subset Counting

Let $A$ be a finite set with cardinality (eleman sayısı) $s(A) = n$. Every element in $A$ has two independent choices when forming a subset: either it is included or it is excluded.

• Total Subsets (Alt Küme Sayısı): $2^n$
• Proper Subsets (Öz Alt Küme Sayısı): $2^n - 1$ (all subsets except the set $A$ itself)
• $k$-Element Subsets ($k$ Elemanlı Alt Küme Sayısı): $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n$ and $\binom{n}{k} = \binom{n}{n-k}$.

Subsets with Element Constraints

Whenever a problem locks the status of $p$ specific elements (either forcing them in or forcing them out), those $p$ elements lose their freedom of choice, leaving $n - p$ free elements:

1. Contains (or does NOT contain) $p$ specific elements: Both yield $2^{n-p}$ subsets. Even if $p_1$ elements must be in and $p_2$ elements must be out, the count is $2^{n - (p_1 + p_2)}$.
2. Contains at least one of $\{a, b\}$ ($a$ veya $b$ bulunur): Subtract the subsets containing neither $a$ nor $b$ from the total:

$$\text{Total} - \text{Neither}(a, b) = 2^n - 2^{n-2}$$

More generally, the number of subsets containing at least one of $p$ specified elements is $2^n - 2^{n-p}$. 3. Contains $a$ OR $b$, but NOT both: Either $a \in S, b \notin S$ ($2^{n-2}$ ways) or $a \notin S, b \in S$ ($2^{n-2}$ ways), giving $2 \cdot 2^{n-2} = 2^{n-1}$ subsets.

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2. Set Operations, Complements & Inclusion-Exclusion

For sets $A, B, C$ inside a universal set $E$ (evrensel küme), the Inclusion-Exclusion Principle (Birleşim Kümesinin Eleman Sayısı) prevents double-counting elements in intersections:

$$s(A \cup B) = s(A) + s(B) - s(A \cap B)$$

Partitioning $A \cup B$ into three mutually disjoint regions (ayrık bölgeler) gives the high-speed alternative:

$$s(A \cup B) = s(A \setminus B) + s(B \setminus A) + s(A \cap B)$$

For three sets:

$$s(A \cup B \cup C) = s(A) + s(B) + s(C) - s(A \cap B) - s(A \cap C) - s(B \cap C) + s(A \cap B \cap C)$$

Complements (Tümleyen) & De Morgan's Laws

The complement $A'$ consists of all elements in $E$ outside $A$:

• Universal Partition: $s(A) + s(A') = s(E)$, which implies $s(A) + s(B') + s(A') + s(B) = 2 \cdot s(E)$.
• Set Difference Conversion: $A \setminus B = A \cap B'$.
• De Morgan's Laws: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$.

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3. Cartesian Products (Kartezyen Çarpım)

The Cartesian product of $A$ and $B$ is the set of all ordered pairs $A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$.

• Cardinality: $s(A \times B) = s(A) \cdot s(B)$, and $s(A \times (B \cup C)) = s(A) \cdot s(B \cup C)$.
• Geometric Enclosing Formulas: When $A, B \subset \mathbb{R}$ are finite sets of numbers, plotting $A \times B$ on the coordinate plane creates a rectangular grid of lattice points with horizontal span $\Delta x_{\max} = \max(A) - \min(A)$ and vertical span $\Delta y_{\max} = \max(B) - \min(B)$:
  • Smallest Enclosing Rectangle Area: $\Delta x_{\max} \cdot \Delta y_{\max}$
  • Smallest Enclosing Circle Radius (En Küçük Çemberin Yarıçapı): The diagonal of the bounding rectangle is the diameter $d = \sqrt{(\Delta x_{\max})^2 + (\Delta y_{\max})^2}$, so the radius is:

$$R = \frac{1}{2}\sqrt{(\Delta x_{\max})^2 + (\Delta y_{\max})^2}$$

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4. Abstract Binary Operations (İşlem)

An abstract binary operation $\star$ on a set $S$ assigns a formula to $x \star y$ (for example, $x \star y = x + y - \frac{1}{2}xy$). Never guess the identity element to be $0$ or $1$—always solve from the axioms in strict order:

1. Identity Element (Birim / Etkisiz Eleman $e$): Set $x \star e = x$ (and $e \star x = x$) and solve for $e$ so that $e$ is independent of $x$.
2. Inverse of an Element ($x$'in Tersi $x^{-1}$): Using the exact number $e$ found in Step 1, set:

$$x \star x^{-1} = e$$

and solve for $x^{-1}$. Check $x^{-1} \star x=e$ as well; the two conditions coincide for a commutative operation. 3. Absorbing Element (Yutan Eleman $y$): An element $y$ is absorbing if $x \star y=y$ and $y \star x=y$ for every $x$ in the set. For the specific commutative operation $x \star y=x+y-\frac12xy$, the identity is $0$ and $2$ is absorbing: $x \star 2=x+2-x=2$. Since $2 \star z$ always equals $2$, it cannot equal the identity $0$, so $2$ has no inverse. Do not infer an absorbing element merely from a vanishing factor in an identity equation; substitute it into both defining conditions.

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5. Modular Arithmetic (Modüler Aritmetik)

For integers $a, b$ and modulus $m \ge 2$, we write $a \equiv b \pmod m$ if and only if $m$ divides $(a - b)$, meaning $a$ and $b$ leave the same remainder when divided by $m$.

• Algebraic Compatibility: If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a \pm c \equiv b \pm d \pmod m$, $a \cdot c \equiv b \cdot d \pmod m$, and $a^n \equiv b^n \pmod m$.
• Cyclic Power Reduction: To evaluate $a^N \pmod m$ for $N\ge0$ when $\gcd(a,m)=1$:
  1. Reduce the base modulo $m$: replace $a$ with $r \equiv a \pmod m$ (using negative remainders like $m - 1 \equiv -1 \pmod m$ whenever convenient).
  2. Compute successive powers $r^1, r^2, r^3, \dots \pmod m$ until you find the smallest positive period $k$ such that $r^k \equiv 1 \pmod m$ (by Fermat's Little Theorem, if $p$ is prime and $p \nmid a$, then $a^{p-1} \equiv 1 \pmod p$, so the period $k$ divides $p - 1$).
  3. Divide the exponent $N$ by $k$: if $N = k \cdot q + s$ with $0 \le s < k$, then:

$$a^N \equiv (r^k)^q \cdot r^s \equiv 1^q \cdot r^s \equiv r^s \pmod m$$

• Units Digit (Birler Basamağı) Problems: Asking for the units digit of a number is identical to computing that number $\pmod{10}$. For instance, to find the units digit of $2027^{2026}$, reduce the base $2027 \equiv 7 \pmod{10}$. The powers of $7$ modulo $10$ cycle with period $k = 4$: $7^1 \equiv 7$, $7^2 \equiv 9$, $7^3 \equiv 3$, $7^4 \equiv 1 \pmod{10}$. Dividing the exponent $2026$ by $4$ gives remainder $2$ ($2026 = 4 \cdot 506 + 2$), so $2027^{2026} \equiv 7^2 \equiv 9 \pmod{10}$.

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Summary Reference Table

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Worked Examples

Worked Example 1: Constrained Subsets & Minimum Enclosing Circle of $A \times B$

Problem: Let $A = \{-3, -1, 0, 2, 4, 5, 7\}$ and $B = \{-2, 1, 6\}$.

1. How many subsets of $A$ contain element $0$, do not contain element $7$, and contain at least one of the negative elements $\{-3, -1\}$?
2. What is the radius of the smallest circle in the coordinate plane that encloses all points of $A \times B$?

Solution:

1. Here $s(A) = 7$. First, lock the two mandatory conditions: $0$ must be included and $7$ must be excluded. That removes $2$ elements from consideration, leaving $7 - 2 = 5$ free elements $\{-3, -1, 2, 4, 5\}$, which generate $2^5 = 32$ subsets. From these $32$ subsets, subtract those that contain neither $-3$ nor $-1$. Locking $-3$ and $-1$ out leaves only $\{2, 4, 5\}$ ($3$ free elements), or $2^3 = 8$ subsets. Thus, $32 - 8 = \mathbf{24}$ subsets satisfy all constraints.
2. For $A \times B$, the horizontal span is $\Delta x_{\max} = 7 - (-3) = 10$ and the vertical span is $\Delta y_{\max} = 6 - (-2) = 8$. The diameter of the smallest enclosing circle is $d = \sqrt{10^2 + 8^2} = \sqrt{164} = 2\sqrt{41}$, so its radius is $R = \frac{d}{2} = \mathbf{\sqrt{41}}$.

Worked Example 2: Abstract Binary Operation & Modular Exponent Reduction

Problem:

1. On $\mathbb{R}$, define $x \star y = x + y - \frac{1}{2}xy$. Find the identity element $e$, the absorbing element $y$, and the inverse $6^{-1}$.
2. Find the remainder when $2027^{2026} + 19^{103}$ is divided by $10$.

Solution:

1. Set $x \star e = x$:

$$x + e - \frac{1}{2}xe = x \implies e\left(1 - \frac{x}{2}\right) = 0 \implies e = \mathbf{0}$$

The factor $1 - \frac{x}{2}$ vanishes when $x = \mathbf{2}$, which is the absorbing element (indeed, $2 \star y = 2 + y - y = 2$ for every $y$). Now find $6^{-1}$ using $6 \star 6^{-1} = e = 0$:

$$6 + 6^{-1} - \frac{1}{2}(6)(6^{-1}) = 0 \implies 6 - 2(6^{-1}) = 0 \implies 6^{-1} = \mathbf{3}$$

1. Modulo $10$, we already know $2027^{2026} \equiv 7^{4(506)+2} \equiv 7^2 \equiv 9 \pmod{10}$. For the second term, $19 \equiv -1 \pmod{10}$, so $19^{103} \equiv (-1)^{103} = -1 \pmod{10}$. Adding the two gives $9 + (-1) \equiv \mathbf{8} \pmod{10}$.

If the base and modulus are not coprime, a power need not return to 1. For example $2^N$ modulo 8 is 0 for all $N\ge3$. Track the initial nonrepeating terms before any eventual cycle. For a noncommutative operation, identity, inverse and absorbing conditions must hold on both sides. An identity-row factor vanishing does not by itself prove absorption; substitute the proposed element into the operation and check the defining conditions.

Review the Method

In modular arithmetic, large exponents are reduced by finding the smallest positive period k where a^k ≡ 1 (mod m), provided gcd(a,m)=1 and replacing the exponent with its remainder modulo k.