can i prove 2 + 2 = 5 mathematiically?
493260
Knowing what a proof is, is the first step in answering this question. From what I am familiar with...\n\nA proof is: a set of logical steps acquired through deductive (therefore, not making any giant leaps in logic, unless by definition), and hence, empirically (from the evidence provided) resulting in a direct equivalence (being, among other types of equivalence, but primarily, in permutation,  multiplicative/additively & negatively/positively & even/odd... meta-mathematically) of states, that's shortest distance is (in absolute terms), either infinity, zero, and/or, also, one.\n\nReally, the attempted 'proof' of 2 + 2 = 5 is based on a distorted type of Trigonometry, which was in essence the source of today's Calculus (just try to draw Tangent or Secant without running into the idea of Calculus' derivative & integral, respectively), and actually is the result of any additive equavalence of any two numbers' to being alike to any number, (because measuring hypotenuse of a given sides is essentially multiplicative, hence partially irrational). \n\n(Which makes me wonder... is there a 2 * 2 = 5 equivalent? and the answer is a resounding, yes! But first the 'proof' as written by Charles Seife.)\n\nLet a & b each be equal to 1. Since a ^ b are equal,\n\nb^2 = ab ...(eq.1)\n\nSince a equals itself, it is obvious that \n\na^2 = a^2  ...(eq.2)\n\nSubtract equation 1 from equation 2. This yeilds\n\n(a^2) - (b^2) = (a^2)-ab ...(eq. 3)\n\nWe can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get\n\n(a+b)(a-b) = a (a-b) ...(eq.5)\n\nSo far, so good. Now divide both sides of the equation by (a-b) and we get\n\na + b = a  ...(eq.5)\n\nb = 0 ...(eq.6)\n\nBut we set b to 1 at the very beginning of this proof, so this means that\n\n1 = 0 ...(eq.7)\n\n ...Anyways, getting that far gives us the jist of the proof, later in the proof, Charles Seife goes on to prove that Winston Churchill was a carrot! if you want to know how that is possible, I recommend you read the book.\n\nFrom equation 7, add a number to either side and get it equal to any other number, one greater than itself.\n\nMultiplying equation 7 after adding to it, and one can get: any number is equal to any other number. \n\nHence, conceptually, any number is equal to zero, and, theoretically, that includes infinity. But that's also the reason why when you divide by zero, it is 'Undefined.' Which, consequentially, is what is happening in this equation... just subsistute 1 into equation 3 and one will see that we are dividing by zero in equation 5. \n\nThis is what lead to the invention of calculus. Really, from here this segways into Hilbert Space... but that is best left for another entry, hopefully, on the actual subject of quantazation.\n\nThat's all I have time for... \n\nTHIS PROOF IS BY DEFINITION INCORRECT, but it provides a good tool as of why we define things in mathematics the way we do.\n\nA good question to ask from here would be (based on my previous tangent):\n\nDoes 1/3 plus 1/3 plus 1/3 = 1?\nOr, does it equal just zero point nine repeating?
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
166772
2x - 3y = 5        \n2x^2 + 3y^2 = 35 \n\n2x = 5 + 3y\nx = (5 + 3y)/2\n\n\n2 [(5 + 3y)/2]^2 = 5 + 3y\n2 [(25+30y+9y^2)/4] = 5 + 3y\n25+30y+9y^2 = 10+6y\n9y^2-24y+15=0\n(3y-3)(3y-5)=0\n\ny=1; y=5/3\n\nif y=1\n2x - 3(1) = 5\nx = 4\n\nif y = 5/3\n2x - 3(5/3) = 5\nx = 5
1023705
2 + 2 / 2 =\n2 + 1 = 3
19885
(5) x (2x^2-3x+5) I think
18182
2x^2 + 5x + 20x + 50\n2x^2 + 25x + 50
584167
7+3+5=15\n6+7+2=15\n8+2+5=15
336800
let a = b\na² = ab (Multiply both sides by a)\na² + a² - 2ab = ab + a² - 2ab (Add {a² - 2ab} to both sides)\n2(a² - ab) = a² - ab (Factor the left, and collect like terms on the right)\n2 = 1 (Divide both sides by {a² - ab})\n\nOnce you've got that settled, you can start on 2 + 2 = 5:\n\n2 = 1 + 1, so therefore\n\nx = 2 + 2 goes to\n\nx = 2 + (1 + 1)\n\nBecause you now have proof that 1 = 2\n\nx = 2 + (1 + 2)\n\nx = 2 + 3\n\nx = 5\n\nSubstituting the original equation for x:\n\n2 + 2 = 5\n\nBut also, since 1=2 and 2=1 then 1+1=4 and 2+2=2
988599
2^1999 x 5^2000 = \n2^1999 x 5^1999 x 5 =\n(2 x 5)^1999 x 5 =\n5 x 10^1999\nThis is 5 followed by 1999 zeroes.\nThis an exact aresult.
730037
let that number be called x.\n\nso, 'the sum of x and 2/5 of x is 56'\n\nx + 2x/5 (two-fifths of x) = 56\n5x + 2x = 280\n7x = 280\nx = 40
838051
5(x – 2)^2 = 3\n(x – 2)^2 =3/5\nx -2 =+- sqrt(3/5)\nx=2+- sqrt(3/5)
817933
2 + 1 = 3\n4 + 1 = 5\nTake the numerals out of the two equations above which are in the next equation 2+2=5.\nWe will get 2 & 5.Now take out the nos. which are similar in both the equations.now we will have 1,1,2,5.As we have 3 numerals in the final equation we will add up the smallest digits 1+1.Then we will have 2,2,5which also means 2+2=5.\n2+2=5 is equal to 2+1+3and 4+1+5\nA little confusing.\nI hope u understand.
811163
(2 * 20) - 35 + 5 = 10
379711
Easy Homework?\n\na=2, b=1, k=2, j=5, n=3.\n2^2=1^2 (mod 3), and 2=5 (mod 3) but 2^5=2 (mod 3) and 1^5=1 (mod 3), so 2^5 is not 1^5 (mod 3).
904926
5+-V92\nquadratic formula
840880
assuming your expression equals 0\nx^2-25-0\n(x+5)(x-5)=0\nx=5 or -5
120340
While it isn't possible, and all these "proofs" are fake (i.e. they contain a methematical error), I'll provide you one just for fun.\n\nLet a and b be two identical numbers (a = b)\n\na^2 = ab\na^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)\n2(a^2 - ab) = a^2 - ab\n\nDividing both sides by (a^2 - ab) gives you\n\n2 = 1\n\n2 = 1 + 1 (common mathematical knowledge)\n\n2 = 2 + 1 (since 1 = 2)\n2 = 2 + 2 (again)\n1 + 1 = 2 + 2\n2 + 2 = 2 + 2\n1 + 1 + 2 = 2 + 2\n2 + 1 + 2 = 2 + 2\n5 = 2 + 2\n\n2 + 2 = 5