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Difference between revisions of "User:Jiang147369"
 
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Therefore, <math>9+10 = 21</math>
 
Therefore, <math>9+10 = 21</math>
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==Contributions==
 +
Here is a list of my AoPS Wiki contributions.
 +
 +
===1975 AHSME===
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* [[1975_AHSME_Problems/Problem_14|Problem 14]]
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* [[1975_AHSME_Problems/Problem_17|Problem 17]]
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* [[1975_AHSME_Problems/Problem_18|Problem 18]]
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* [[1975_AHSME_Problems/Problem_19|Problem 19]]
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* [[1975_AHSME_Problems/Problem_23|Problem 23]]
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* [[1975_AHSME_Problems/Problem_25|Problem 25]]
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 +
===1976 AHSME===
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* [[1976_AHSME_Problems/Problem_12|Problem 12]]
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* [[1976_AHSME_Problems/Problem_13|Problem 13]]
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* [[1976_AHSME_Problems/Problem_14|Problem 14]]
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* [[1976_AHSME_Problems/Problem_15|Problem 15]]
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* [[1976_AHSME_Problems/Problem_17|Problem 17]]
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* [[1976_AHSME_Problems/Problem_21|Problem 21]]
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* [[1976_AHSME_Problems/Problem_23|Problem 23]]
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* [[1976_AHSME_Problems/Problem_29|Problem 29]]
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===1977 AHSME===
 +
* [[1977_AHSME_Problems/Problem_6|Problem 6]]

Latest revision as of 14:50, 17 August 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

1977 AHSME

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