Facebook math
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Facebook math
https://brilliant.org/communityproblem/aproblembyarulkolla/?group=i2KL1625hqMK
(What is 2^2014  2^2013?)
I didn't want to comment on the facebook link, so I'll write crap here.
This is a series of characters and formatting, presumably to represent a mathematical expression using arabic numerals of base ten. To expand on that: This expression presumably uses elements of the ring integers (We'll call this set Z) under standard addition and multiplication (We'll call this ring ZR). The term  is presumably shorthand for the function :ZR>ZR with ab = a + b * 1 given a,b in ZR. The ^ operator (superscript in the link) is not a ring operator, but, assuming it is standard exponentiation, it is trivial to show ^ is closed for the left operand an integer, and the right operand a positive integer, and therefore should cause no problems with evaluation of the expression. I will also assume we are following the standard notation, which requires operations of ^ to occur before operations of  or + or *. In that case, it is fairly simple to evaluate the expression, assuming it is to be evaluated alone. 2^2014  2^2013 can be rewritten as 2*2^2013  2^2013 or (21)2^2013 using the distributive property of rings. We are also using the ( ) as a notation for precedence of evaluation, so (21)*2^2013 = 1*2^2013. We know 1 is the multiplicative identity of the ZR ring, therefore 2^2014  2^2013 = 2^2013.
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Math class withdrawal sucks
(What is 2^2014  2^2013?)
I didn't want to comment on the facebook link, so I'll write crap here.
This is a series of characters and formatting, presumably to represent a mathematical expression using arabic numerals of base ten. To expand on that: This expression presumably uses elements of the ring integers (We'll call this set Z) under standard addition and multiplication (We'll call this ring ZR). The term  is presumably shorthand for the function :ZR>ZR with ab = a + b * 1 given a,b in ZR. The ^ operator (superscript in the link) is not a ring operator, but, assuming it is standard exponentiation, it is trivial to show ^ is closed for the left operand an integer, and the right operand a positive integer, and therefore should cause no problems with evaluation of the expression. I will also assume we are following the standard notation, which requires operations of ^ to occur before operations of  or + or *. In that case, it is fairly simple to evaluate the expression, assuming it is to be evaluated alone. 2^2014  2^2013 can be rewritten as 2*2^2013  2^2013 or (21)2^2013 using the distributive property of rings. We are also using the ( ) as a notation for precedence of evaluation, so (21)*2^2013 = 1*2^2013. We know 1 is the multiplicative identity of the ZR ring, therefore 2^2014  2^2013 = 2^2013.
/\
\/
Math class withdrawal sucks
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Re: Facebook math
I tried to submit the answer on that website but it said I had to register in order to do so. As if I'm going to give them my personal information so I can do math problems. As if my college classes don't give me enough math problems to do...
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