document.write( "Question 319222: Let symbol ~ be defined so that a~b equals the smaller of a-b and b-a. For all real numbers a and b, all of the following are true EXCEPT:
\n" ); document.write( "a) a~b = b~a
\n" ); document.write( "b) a~b = (-a) ~ (-b)
\n" ); document.write( "c) a~b < 0
\n" ); document.write( "d) a~0 = 0 ~ (-a)
\n" ); document.write( "e) If a>b>0, then (a~b)~b = -a
\n" ); document.write( "The answer key tells me that the correct answer is C, but it seems to me that using any combination of numbers will always result in a negative response.
\n" ); document.write( "Two negatives: -3 - (-5) = 2 *************** -5 - (-3) = -2
\n" ); document.write( "Two positives: 5-4=1 ************ 4-5= -1
\n" ); document.write( "One negative one positive: 5- (-3)= 8 **************** -3 - (5) = -8
\n" ); document.write( "Does 0 count as a real number? Is that why the solution is what it is?
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Algebra.Com's Answer #228569 by jim_thompson5910(35256) $\"\"$ $\"About$

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The statement a~b < 0 means that a ~ b is less than zero for ALL values of 'a' and 'b'. However, if $\"a=b\"$ , then $\"a-b=a-a=0\"$ and $\"b-a=a-a=0\"$ which basically means that $\"a-b=b-a=0\"$ . So if $\"a=b\"$ , then a ~ b = 0 meaning that a ~ b < 0 is NOT true. Does that make sense? \n" ); document.write( "

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