1=2  ture or false? look at proof?
146536
Until ths step 7, it is allright\n2 (a^2 - a b) = 1 (a^2 - a b).\n\nBut you can't divide by (a^2-ab),which equal zero, and it is not allowed to divide by zero\n\na=b \na^2=ab \na^2-ab=0\n\nwhy because 5/0 =6/0 =7/0 =infinty and this doesn't mean that 5=6=7
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
717483
1\nbecause u have mentioned that 1=5 so 5=1 and it cant = 9
317141
To prove that a statement is false, all you have to do is find ONE case where it is false.\nLet a = 1 and b = 1\n(a+b)^3 = (1+1)^3 = 2^3 = 8\na^3+b^3 = 1^3 + 1^3 = 1 + 1 = 2\n8 does not equal 2.\nThus, I have just proven that it is false.
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
739421
x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y
541208
it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
456169
There are many proofs but with fallacies. Please visit:\nhttp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html\n\nhttp://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html
324840
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
471450
weezy is on the right track, that glow1989 is on the right track.  For everyone who thinks that writing 1+1=2 is so obvious that it constitutes a self-contained proof of itself(!)... if you undertake a serious study of mathematical proof, you might find that 357 pages is a good length for a quick introduction to the ideas that are involved in considering questions in this area.  Of course, I am talking about actually UNDERSTANDING something - very different to simply passing an exam, regardless of the 'level' of maths you are enrolled in.\n\nYes, as you mentioned, this is a hard question.  'Proving' that 1+1=2 is not as simple as it sounds, because of the inherent question of what we mean by 'prove'.  You might have noticed that some proofs you have used depend on other proofs, which in turn depend on other proofs, ... which ultimately rest on a handful of axioms.  It is the nature of operations that are allowed by any axiomatic system, that determine the characteristics of mathematical problems that can be said to be SOLVABLE.  It is no small task to understand the axioms that underlie our systems of mathematics.  For example, maths students learn: that division by zero is undefined; that any object raised to a null power is equal to unity; some differences between rational and irrational numbers; what we mean when we consider objects to exist in 'sets'; and so on.  Those who consider themselves to be excellent students are usually able to list these rules in rapid-fire style.  But if we are asked to 'prove' why those rules are the way they are, the questions start to seem fuzzy, and you typically run out of convenient formulas to rely on.  You could easily write 357 pages just to explain what you actually MEAN by the numbers 1 or 2, without worrying about the definition of the addition operation!  As usual, people might do better to stop and consider what the question is asking before booting in a smug answer.\n\nLook into Hilbert's Problem (David Hilbert - German mathematician), which was a more workable and compact form of a proposal by Russell and Whitehead to formalise mathematics on a general basis, according to well-established axioms and procedures.\n\nOther notable mathematicians to work in this area were Alan Turing, Alonzo Church (look for Church's Lambda Calculus), Emil Post, and the brilliant Kurt Gödel.  In more recent decades, look for Gandy (1988) and of course many others.\n\nLol, molawby is spot-on.  I wish he'd posted b4 me... would've saved me a bit of typing.  If all you need to do is learn to sing the right song, to pass your exams and get the job that stipulates that you need to have completed such-and-such a course, then you don't have to worry about any of this.  Of course, that doesn't mean you have to dismiss it either.  But if you are chasing a deeper understanding, it probably does take the 20 yrs, and you won't necessarily get paid for it, heheheh.\n\nFor bedside reading, I suggest Roger Penrose etc...
311472
i can prove that\n' a = x [true for some a's and x's]\na+a = a+x [add a to both sides]\n2a = a+x [a+a = 2a]\n2a-2x = a+x-2x [subtract 2x from both sides]\n2(a-x) = a+x-2x [2a-2x = 2(a-x)]\n2(a-x) = a-x [x-2x = -x]\n2 = 1 [divide both sides by a-x] ' is wrong\nbecause if we assume that a=x, we can't divide by a-x which is 0!!!! there is no way anyone can prove that 2=1!!!! don't even try:)\n\nalso, the one with 2^0 and 1^0\nany number ^0 is 1, does it mean that 1=2=3=4=5=6=7=8=9=...\nalso, for example 2^2=4 but (-2)^2=4 too, does it mean that 2=-2????? NO!!!
249413
1 + 1 = 2(+n) ;not really a function
169166
well i think yoru referring to fermats last theorem. This was proven by Andrew Wiles of Britan in 1994-1995.  He squestered himself in a room for more or less 7 years and came up with a solution.  Originally his proof was rejected when he first presented it in 1993 because an error was found.  He fixed the error and we now know that Fermats last Theorem is true.
930851
... 1 + 1 + 1 equals 3.\n\nDo you make any sense AT ALL?
153034
No, nobody really can prove it, with assuming several axioms\nfirst. There are certain things you have to assume are true in your universe, without necessary being able to check them. From those axioms, you can derive conclusions (theorems) that are always true, but only in that particular\nuniverse. The discussed operation of addition is one such theorem. If you assume that the axioms of natural numbers are true, then this addition becomes quite trivival. If you assume those axioms of natural numbers to be false, then the above operation of addition might become false with a different set of supposition. In other words, only with the facts given, one cannot prove, nor disprove, the fact of addition. It all depends on the set of assumptions you would have made before adding the two ones.
523701
If you add the numbers together then they equal an even number.   1 1  1+1=2
767112
Here's what you should think: 3 times a number that has to be zero or higher, then having another 3 added to it, cannot possibly equal 1 (it must be at least 3).  So there is no solution.\n\nYou did say it was an inequality, though.  If the = was really > or >=, the answer is all real numbers.  If it was < or <= the answer is the empty set.