Disprove 3=2,  please!?
401494
From,\n(3-5/2)^ 2 = (2-5/2)^ 2\n\nThe mistake is in \n"Taking positive square root on both sides"\n\nLHS = Left Hand Side\nRHS = Right Hand Side\n\nFor we cannot simply take square roots. \nLook at LHS (3-5/2) is the POSITIVE root. \nOn the RHS (2-5/2) is the NEGATIVE root.\nYou made them "equal" to get the "wrong proof".\n\nActually, to take the roots note that \n(2-5/2)^ 2 = (5/2 - 2)^ 2\nSo, If you wanted to take "positive root" then,\nits obviously (5/2 - 2).\n\nThus comparing positive roots,\n3 - 5/2 = 5/2 - 2 which is a identity.\n\nRegards
161410
Okay.  I'm explaining this to one of the people above me, too.\n\n1/3 = .33\n2/3 = .66\n\nTherefore, if logic persists, 3/3 = .99.  However, this is not true because math is dumb like that.  \n\nIf that concept was true, odd things would happen in the math world.  Basically, if the numbers keep continuing forever, you're never going to have the next number, so you never could have even began.  You would be stuck in NUMBER LIMBO.  There would BE NO NUMBERS.  And everything would be pretty much screwed.\n\nThis is why 3/3 = 1.  But also, 4/3 is obviously more than one.  It's all very confusing, and there are loopholes everywhere.  This is why people round and I am not in charge of math books.
910422
no \n3x-1>1 \n3x>2 \nx>2/3
794793
x = 3. Try entering it on your calculator.
1123927
two cant equal three because two is two and three is three.  And by sloving, it would mean that you are solving for the X. Which is zero.  If you every, in any equation end with 2=3 or something like that, it is considered 'no solution'
2302
(1) = x\n(2) = / \n(3) = x
321226
It's false.\n\nYou have to use order of operations. You do the stuff in the parenthesis first:\n\n(3-3) - 3 = 3- (3-3)\n(0)-3=3-(0)\n\nSo you get:\n-3=3\n\nSince -3 isn't equal to 3, it's not true.
56274
In a base 3 group, 1+2=10 because there are only 3 digits to express any amount: 0, 1 and 2.\n\nMaybe not the answer you are looking for, but, what the heck, in math terms is completely correct :D
203409
k(k+2)! = k.1.2...k.(k+1)(k+2)\n(k+3)! = (k+3)(k+2)! \n=>\nk/(k+3) i think that's enough. that's true if k is a Natural number, there might be some theoremes that say that's true when k is Real but I'm not sure
884616
Possible .X....Y\noutcome\nHHHH......3....3\nHHHT......3....2\nHHTH......2....2\nHTHH......2....2\nHHTT......2....1\nHTHT......2....1\nHTTH......1....1\nHTTT......1....0\nTHHH......2....3\nTHHT......2....2\nTHTH......1....2\nTTHH......1....2\nTHTT......1....1\nTTHT......1....1\nTTTH......0....1\nTTTT......0....0\n\nHence,\nP(X=3,Y=3)=1/16\nP(X=3,Y=2)=1/16\nP(X=3,Y=1)=0\nP(X=3,Y=0)=0\nP(X=2,Y=3)=1/16\nP(X=2,Y=2)=3/16\nP(X=2,Y=1)=2/16\nP(X=2,Y=0)=0\nP(X=1,Y=3)=0\nP(X=1,Y=2)=2/16\nP(X=1,Y=1)=3/16\nP(X=1,Y=0)=1/16\nP(X=0,Y=3)=0\nP(X=0,Y=2)=0\nP(X=0,Y=1)=1/16\nP(X=0,Y=0)=1/16\n\n.................JOINT PROBABILITY DISTRIBUTION OF x & y\n\n........0.1....2.3.....|P(X=x)   \n...&#92;Y|__________|\n..X&#92;.|..........................|\n..__|..........................|\n....0|1/16.1/16.0.0.....|2/16\n......|..........................|\n....1|1/16.3/16.2/16.0|2/16\n......|..........................|\n....2|0.2/16.3/16.1/16|6/16\n......|..........................|\n....3|0....0..1/16..1/16|2/16\n......|..........................|\nP(Y=y)|2/16.6/16.6/16.2/16|1\n\nfOR 2nd part we have to find \nP(X=3|Y=1)=P(X=3,Y=1)/P(Y=1)=0\nP(X=2|Y=1)=P(X=2,Y=1)/P(Y=1)=(2/16)/(6/16)=1/3\nP(X=1|Y=1)=P(X=1,Y=1)/P(Y=1)=(3/16)/(6/16)=1/2\nP(X=0|Y=1)=P(X=0,Y=1)/P(Y=1)=(1/16)/(6/16)=1/6\nHence\n      conditonal probability distritribution of X given Y=1\n......X...|0.1.2.3..|TOTAL\n.....___|_______|_____\nP(X=x)|0.1/3.1/2.1/6.|1\n\nForget about those (.........)s. They are provided so that the tables do not collapse.
218316
Thats funny.  Tell me another one.\n\nMaybe its like 5-2=2\n\n\nlol
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
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
816271
Assuming each is equal to y, the domain and range of the first two are all real numbers, since there is no x in a denominator, square root, etc.\n\nHowever the third cannot have x = 1/2 [if the whole 2x-1 is the denominator].  If only the 2 is in the denominator the domain and range are all numbers; if only the 2x is in the denom. the domain is all numbers but 0.  In case 1 , the range is all numbers but zero since a fraction with 1 on the top can never equal zero.  In case 3, the range is all numbers but -1 for a similar reason.  Good luck!
890832
Gary, that's a good one, but one mistake...\nyou said that a=b, thus you cannot divide by (a-b) b/c that equals zero, and you cant divide by zero.
269006
Ask your friends if they can find the error in the following proof.\n(it's obviously not right, but it looks cool!)\n\na=b\na+1 = b+1\n(a+1)^2 = (b+1)^2\na^2 + 2a +1 = b^2 + 2b +1\ncancel the ones and rearrange\na^2 - b^2 = 2(b-a)\n(a+b)(a-b) = 2(b-a)\ntherefore a+b = -2\nPick numbers for a and b\nand look!! 5 + 7 = -2\nThus shown.\nLOL
958343
you evil Troll! You asked a trick question!\n\n(half of 2) plus 2=3\n            -BUT-\nhalf of (2 plus 2)=2