can anyone prove 1+1=3?
610821
Easy...\n0 = 0\n\nadding -2 on both sides\n\n-2 = -2\n-> -3 + 1 = -6 + 4\n-> -2*3/2 + 1 = -4*3/2 + 4\n\nadding 9/4 on both sides..\n\n-> 9/4 -2*3/2 + 1 =9/4 -4*3/2 + 4\n->\n(3/2 - 1)^2 = (3/2 - 2)^2\n\ntaking +ve sqr root\n3/2 - 1 = 3/2 - 2\n\n-> 1= 2\nadding 1 on both sides\n2 = 3\n-> 1 + 1 =3
919554
Factor out 3.\n3(x^2+2x+1)=0\n3(x+1)^2=0\nThe only answer then is x=-1 to make the expression equal 0.
287019
1 + 1 = 2\none plus one = oneone
369170
same as everywhere\n1+1=2\n2+2=4
1023705
2 + 2 / 2 =\n2 + 1 = 3
1092372
6\n1+(1*2)+3\n1+2+3\n3+3\n6\n\nuse order of opperations.  PEMDAS
1109416
3/x  + 1    the way you put it..
57681
1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1\n\nYou can prove it by induction:\nFor n=1, 1(1!) = 2!-1 = 1\nNow assume 1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1 for n. Then show that it is true for n+1 as well: \n\n1(1!) + 2(2!) + ... n(n!) = (n+1)!-1\nAdd (n+1)(n+1)! to both sides:\n1(1!) + 2(2!) + ... n(n!) + (n+1)(n+1)! \n= (n+1)!-1 + (n+1)(n+1)!\n= (n+1)!(1+(n+1)) - 1\n= (n+1)!(n+2) - 1\n= (n+2)! - 1\n\nSince the equation is true for n=1, and when it is true for n it is also true for n+1, it must be true for all positive integers.
1157222
You know there are 8 axioms in mathematics which we cannot prove\n\n1) if 'a' and 'b' are integers then 'a+b' is an integer and 'a.b' is an integer (Closure law)\n2) a + b = b + a and a..b = b.a (cumulative law)\n3) a + (b + c) = (a + b) + c (Associative law)\n4) a (b + c) = a.b + a.c (Distributive law)\n5) a + 0 = a = 0 + a (Additive Identity)\n6) a.1 = a = 1.a (Multiplicitive Identity)\n7) for all integer 'a' there is a unique integer (-a) such that a + (-a) = 0\n8) I can't remember...\n\nSo we can prove your requirement using above axioms, \n\na + (-a) = 0\n(-1)(a + (-a) = -1(0)  (Multiplying by same factor)\n-a + -(-a) = 0 (Distributive law over addition)\n\nadding +a both sides\n\na + (-a + -(-a)) = a + 0 \n(a+ (-a)) + -(-a) = a (Associative law)\n0 + -(-a) = a \n-(-a) = a (proved... huh...)
1014029
1,2,3,5,16,63, ...\n3=(1+2)*1\n5=(2+3)*1\n16=(3+5)*2\n63=(5+16)*3
249413
1 + 1 = 2(+n) ;not really a function
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.
78524
(1-a)+3(a-2)-(2a-3)=1
1110742
12 simply place the 1 where the x is\n\n(1+3)(1+2)\n4*3\n12
2302
(1) = x\n(2) = / \n(3) = x
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.
328795
1 + 1 = whatever you want to make.