diff --git a/.obsidian/workspace.json b/.obsidian/workspace.json
index 564c4ce..5457bd9 100644
--- a/.obsidian/workspace.json
+++ b/.obsidian/workspace.json
@@ -158,6 +158,7 @@
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   "active": "81e5aff710e811bc",
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     "UNB/Year 4/Semester 2/CS2333/2024-01-22.md",
     "UNB/Year 4/Semester 2/CS3873/2024-01-22.md",
     "UNB/Year 4/Semester 2/CS3873/2024-01-19.md",
@@ -190,7 +191,6 @@
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     "Semester 1/CS3418/10-27-2023.md",
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     "Semester 1/CS2418",
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diff --git a/UNB/Year 4/Semester 2/CS2333/2024-01-22.md b/UNB/Year 4/Semester 2/CS2333/2024-01-22.md
index e69de29..40c375d 100644
--- a/UNB/Year 4/Semester 2/CS2333/2024-01-22.md	
+++ b/UNB/Year 4/Semester 2/CS2333/2024-01-22.md	
@@ -0,0 +1,76 @@
+Lecture Topic: Relation examples
+
+# Relation Example 1
+$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$
+e.g $(22,7) \in R$ because $22-7 = 5(3)$
+e.g $(7,22) \in R$ because $7-22 = -15 = 5(-3)$
+e.g $(22,9) \notin R$ because $22-9 = 13$, which is not a multiple of 5
+
+Reflexive? Yes
+For every integer a, $a-a = 0 = 5(0)$
+Therefor $(a,a) \in R$ for every integer a
+
+Symmetric? Yes
+Let a, b be any integer, such that (a,b) is in the relation (can we prove that (b,a) is in relation as well?)
+
+Since (a,b) is in the relation, $a-b = 5n$, so, $b-a = -5(n) = 5(-n)$
+Note: $(b-a) = -(a-b)$
+
+Transitive? Yes
+Let a,b,c be any integers such that $(a,b) \in R$ and $(b,c) \in R$
+Can we prove that $(a,c) \in R$ ?
+
+Since $(a,b) \in R$, we know $a-b=5n$ for some $n \in \mathbb{Z}$
+Since $(b,c) \in R$, we know $b-c=5p$ for some $n \in \mathbb{Z}$
+
+Now, $(a-c) = (a-b) + (b-c) = 5n + 5p = 5(np)$
+Therefore, $(a-c) \in R$
+Note: This is an integer because it is the sum of integers
+
+# Relation Example 2
+$R = \{(i,j) | i \in \mathbb{R}, j \in \mathbb{R}, i + 2 > j\}$
+
+Examples: 
+- $(5,2) \in R$ because $(5+2) > 2$
+- $(3,4) \in R$ because $(3+2) > 4$
+- $(0,7) \notin R$ because $(0+2) \ngtr 7$
+
+Reflexive? Yes
+Symmetric? No
+Transitive? No
+
+Proof: We need to find real numbers where $a,b,c$ where $(a,b) \in R$ and $(b,c) \in R$ but $(a,c) \notin R$
+
+For example, a = 1, b = 2, c = 3
+$(1,2) \in R$ because $1+2 > 2$
+$(2,3) \in R$ because $2+2 > 3$
+
+But $(1,3) \notin R$ because $(1+2) \ngtr 3$
+$\therefore$ R is not transitive
+
+# Equivalence relations
+Equivalence relations have all three of these properties
+Any equivalence relation R on a set A induces a partition of A
+- Splits A into equivalence classes
+	- Each class contains elements that are related to themselves and to each other, but to nothing outside the class
+
+So, for the above relation:
+$$R = \{(i,j) | i \in \mathbb{Z}, j \in \mathbb{Z}, i-j =5n \text{ for some integer n}\}$$
+This forms an equivalence relation
+
+So for the set of all integers, this forms 5 equivalence classes:
+- $..., 0, 5, 10, ...$
+- $..., 1, 6, 11, ...$
+- $..., 2, 7, 12, ...$
+- $..., 3, 8, 13, ...$
+- $..., 4, 9, 14, ...$
+
+## Example
+$A = \{-3, -2, -1, 0, 1, 2, 3\}$
+$R = \{(i,j) | i \in \mathbb{A}, j \in \mathbb{A}, i^2 = j^2\}$
+
+So the equivalence classes induced by this relation:
+- -3, 3
+- -2, 2
+- -1, 1
+- 0
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