|
|
| (One intermediate revision by the same user not shown) |
| Line 3: |
Line 3: |
| | (A) 19 (B) 20 (C) 21 (D) 22 (E) 23 | | (A) 19 (B) 20 (C) 21 (D) 22 (E) 23 |
| | | | |
| − | ==Solution 1== | + | ==Solution== |
| − | Define = satisfying the following axioms
| + | By basic math, <math>9+10=\boxed{(\textbf{A}) 19}</math>. |
| − | | |
| − | <math>a=a</math>
| |
| − | | |
| − | <math>a=b \implies b=a</math>
| |
| − | | |
| − | <math>a=b, b=c \implies a=c</math>
| |
| − | | |
| − | Define <math>\mathbb{N}</math>
| |
| − | | |
| − | <math>0 = \emptyset = \{ \}</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)
| |
| − | | |
| − | <math>S(n) := n \cup \{n \}</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>
| |
| − | | |
| − | Define +
| |
| − | | |
| − | <math>a+0=a \forall a \in \mathbb{N}_0</math>
| |
| − | | |
| − | <math>a+S(b)=S(a+b) \forall a,b \in \mathbb{N}_0</math>
| |
| − | | |
| − | <math>a+b=b+a</math>
| |
| − | | |
| − | Name the numbers
| |
| − | | |
| − | <math>1 := S(0) = \{ 0 \} </math>
| |
| − | | |
| − | <math>2 := S(0) = \{ 0, 1 \} </math>
| |
| − | | |
| − | <math>\vdots</math>
| |
| − | | |
| − | <math>21 := S(18)</math>
| |
| − | | |
| − | Now solving 9+10
| |
| − | | |
| − | <math>9+10=9+S(S(S(S(S(S(S(S(S(S(0))))))))))</math> | |
| − | | |
| − | <math>=S(9+S(S(S(S(S(S(S(S(S(0))))))))))</math>
| |
| − | | |
| − | <math>=S(S(9+S(S(S(S(S(S(S(S(0))))))))))</math>
| |
| − | | |
| − | <math>\vdots</math>
| |
| − | | |
| − | <math>=S(S(S(S(S(S(S(S(S(S(9+0))))))))))=S(S(S(S(S(S(S(S(S(S(9))))))))))</math>
| |
| − | | |
| − | <math>=S(18)=21</math>
| |
| − | | |
| − | Therefore 9+10=21 <math>\textbf{(C)} 21</math>
| |
.
