Difference between revisions of "2024 AMC 10A Problems/Problem 14"

(Solution 1)
Line 5: Line 5:
 
==Solution 1==
 
==Solution 1==
 
Define = satisfying the following axioms
 
Define = satisfying the following axioms
<math>a=a</math>
+
<math>a=a</math>\
<math>a=b \implies b=a</math>
+
<math>a=b \implies b=a</math>\
<math>a=b, b=c \implies a=c</math>
+
<math>a=b, b=c \implies a=c</math>\
Define <math>\mathbb{N}</math>
+
Define <math>\mathbb{N}</math>\
<math>0 = \emptyset = \{ \}</math>
+
<math>0 = \emptyset = \{ \}</math>\
<math>0 \in \mathbb{N}_0</math>
+
<math>0 \in \mathbb{N}_0</math>\
(note we use <math>\mathbb{N}_0</math> cause I'm one of those <math>0 \notin \mathbb{N}</math> people)
+
(note we use <math>\mathbb{N}_0</math> cause I'm one of those <math>0 \notin \mathbb{N}</math> people)\
<math>S(n) := n \cup \{n \}</math>
+
<math>S(n) := n \cup \{n \}</math>\
<math>n \in \mathbb{N}_0 \implies S(n) \in \mathbb{N}_0</math>
+
<math>n \in \mathbb{N}_0 \implies S(n) \in \mathbb{N}_0</math>\|
<math>\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n</math>
+
<math>\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n</math>\

Revision as of 12:37, 11 August 2024

Since you came this far already, here's a math problem for you to try: What is 9+10? (A) 19 (B) 20 (C) 21 (D) 22 (E) 23

Solution 1

Define = satisfying the following axioms $a=a$\ $a=b \implies b=a$\ $a=b, b=c \implies a=c$\ Define $\mathbb{N}$\ $0 = \emptyset = \{ \}$\ $0 \in \mathbb{N}_0$\ (note we use $\mathbb{N}_0$ cause I'm one of those $0 \notin \mathbb{N}$ people)\ $S(n) := n \cup \{n \}$\ $n \in \mathbb{N}_0 \implies S(n) \in \mathbb{N}_0$\| $\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n$\