Nooblet94
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- 2012
Re: 2012 HSC MX2 Marathon
For the Fibonnacci/triangle question, is this right? From what I can see it is, but I've never actually done a proof by contradiction before so I might've fucked up somewhere.
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Fibonacci numbers are defined by $F_n=F_{n-1}@plus;F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B@plus;C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B@plus;C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." title="\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." /></a>
Hmmm okay. I'll try it tomorrow after my goddamn legal assessment (which I possibly might fail).
For the Fibonnacci/triangle question, is this right? From what I can see it is, but I've never actually done a proof by contradiction before so I might've fucked up somewhere.
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Fibonacci numbers are defined by $F_n=F_{n-1}@plus;F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B@plus;C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B@plus;C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." title="\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." /></a>
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