Give a counterexample to the following:?
360700
a = 1, b = 2, n = 3:\n\n1^2 = 1 = 4 = 2^2 (mod 3) but clearly 1 != 2 (mod 3)
993026
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
678158
because a+b-c = 0, \nu cant cancel out (a+b-c) from both sides \nu can only cancel if (a+b-c) is not zero, which is not the case. \n\nhint: i should get best answer
609321
One is the inverse of the other
309248
Simple answer.\n\n$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
388398
i think the false statement is C. as the degree of any number is always 1. i dont understand the statement b tho :S
408554
yes. take out an x \nyou answer should be x(x+3)
388704
it cannot be f(x) = a for all x, unless a=0; otherwise you would be pretending that a = a+a with a non-0 !! This doesn't happen in R.
667791
Do you mean "not equal to"?\n<>\nThat's also still supported in some newer languages, although != is more common nowadays.
144290
This is a better link that the previous one posted here!! try it! \n\nhttp://pubs.logicalexpressions.com/Pub0009/LPMArticle.asp?ID=305
616457
of the answers given, only supersede meets the condition.\nThere does not appear to be another.
932028
Yes.\n\nThe argument is valid but not perfectly worded.  \n\nHere A, B, C can be variables as others have pointed out or they are statements that are fulfilled in direct progression.  In fact this is a form of the "Direct Proof"\n\nDirect Proof example:\n\nAs an example we can claim that if we have A:={the addition of two odd numbers x, y |  x+y} it will become an even number which can be statement C.\n\nSo by our knowledge that odd numbers can be written as \nx = 2m+1   and y = 2n+1  where m, n are any integers we can use that as statement B.  Hence B:={x = 2m+1   and y = 2n+1  where m, n are any integers}\n\nDoing a little algebra we have \nx+y = (2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)\n\nNow we can find out that (m+n+1) is an integer as by definition and we can call it "p" s.t. p=(m+n+1),\nHence x+y = 2(m+n+1) = 2p, and by our knowledge of numbers 2p is an even number.  Hence C is fullfilled.\n\nThus like questioner's argument:\nIf A then B then C\nA \ntherefore C.\n\nAnd with my example and statements in quotations can be be replaced by letters A, B, and C: \nIf "you add two odd numbers" then "they are integers that can be written as the addition of two times another integer plus one which algebraically can be written as two times integer" then "you obtain a number which is even".\n"You add two odd numbers" \ntherefore "you obtain a number which is even".\n                                              Q.E.D.
43631
The first part  is (c) either true or false.\nThe second part is also (c)
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
903779
Infinitesimally small.\n\nZero.
902349
Hynton has it.  Just remember that you'll be restricted to x > 0 because of the original form of the problem.