Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "User:Jiang147369"

(→‎1976 AHSME)
(→‎1976 AHSME)
Line 73: Line 73:
 
* [[1976_AHSME_Problems/Problem_13|Problem 13]]
 
* [[1976_AHSME_Problems/Problem_13|Problem 13]]
 
* [[1976_AHSME_Problems/Problem_14|Problem 14]]
 
* [[1976_AHSME_Problems/Problem_14|Problem 14]]
 +
* [[1976_AHSME_Problems/Problem_15|Problem 15]]
 
* [[1976_AHSME_Problems/Problem_17|Problem 17]]
 
* [[1976_AHSME_Problems/Problem_17|Problem 17]]
 
* [[1976_AHSME_Problems/Problem_21|Problem 21]]
 
* [[1976_AHSME_Problems/Problem_21|Problem 21]]
 
* [[1976_AHSME_Problems/Problem_29|Problem 29]]
 
* [[1976_AHSME_Problems/Problem_29|Problem 29]]

Revision as of 13:47, 20 July 2021

Welcome to my page!


Proof of 9 + 10 = 21

Credits to John Hush for the proof of $0 = 1$ and the proof of $1 = 2$


Prove: $0 = 1$

Proof:

Left side $= 0 + 0 + 0 + 0 + 0 + ...$

$= (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + ...$

$= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...$

$= 1$

Therefore, $0 = 1$.


Prove: $1 = 2$

Proof:

Let $a = b$

$a^2 = ab$

$a^2 - b^2 = ab - b^2$

$(a-b)(a+b) = b(a-b)$

$a+b = b$

$b+b = b$

$2b=b$

$2=1$

Therefore, $1 = 2$.


By the Transitive Property, $0 = 2$

Prove: $9 + 10 = 21$

Proof:

$9 + 10 = 19$

$9 + 10 = 19 + 0$

$9 + 10 = 19 + 2$

Therefore, $9+10 = 21$


Contributions

Here is a list of my AoPS Wiki contributions.

1975 AHSME

1976 AHSME

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=User:Jiang147369&oldid=158791"