Difference between revisions of "2024 AMC 10A Problems/Problem 14"
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==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 13:37, 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
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Define 
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(note we use 
cause I'm one of those 
people)\\
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