Prove to me that two (2) plus (+) two (2) equals (=) four (4).?
538579
One starts with the basic concept of unity: 1\n\nOne then defines the successor of 1: it is called 2.\n\nOne then defines addition-of-one:  Anything plus 1 is the successor of that anything; declare that "successor" is commutative in this definition\n\nNEXT...\n\nDefine successor of two as being named 3.\n\nDefine successor of three as being 4.\n\nBUT: 4 is 3 + 1\n\nAND: since 3 is 2 + 1,\n\nTHEN: 4 = (2+1)+1 \n\nBUT then, remembering that 2 is the name for successor-of-1\nand using commutation,\n\n4 = (2+1) +1 = 2+(1+1) = 2 + 2\n\nSO, 2+2 = 4\n\n....  Actually had to do this as a homework problem in an advanced Math course, along about 1972 or so.  I actually had to prove it commuted, but I don't recall how I did that.\n\nObnoxiously simple, really, once you realize that 4 is just the name of successor-of-successor-of-successor-of-1, and all that jazz...
238690
Let S=1+2+3+4+5+...+(n-2)+(n-1)+n\nThen it's the same as S=n+(n-1)+(n-2)+...+3+2+1.\nAddding these two formulas you'll get\n2S=(n+1)+(n+1)+(n+1)+...+(n+1).\nSince there are n terms on the right side,\n we conclude that \n2S=n(n+1) or\n\nS=n(n+1)/2.\n\nHence, in your case it will be\nS=1+2+3+...+966=[966*967]/2=467061.
714877
b)\n\nNote that 314159266 is of the form 4k + 2, where k is an integer.  Since any integer is of the form 2n or 2n +1, this means that all squares of integers are either of the form 4m or 4m + 1.  No matter how we try to do the subtraction of two numbers of the form 4m or 4m + 1, we won't get a number of the form 4k + 2:\n\n4m - 4n = 4k\n4m - (4n + 1) = 4k + 3\n(4m + 1) - 4n = 4k + 1\n(4m + 1) - (4n + 1) = 4k.\n\nCheers!
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
93641
Okay, 2 + 4= Look at 2 now look at 4, 4 is 4 pieces and 2 is 2 pieces. add the 4th piece and the second piece = 6! Duh!
447627
a + 4a where a = 2 ?\n\nif so, then...\n\n2 + (4)2\n\n2 + 8\n\n10\n\nThe answer is 10.
784622
16x^2 - 4x - 2\n= (4x)^2 -(4x)- 2\n= {(4x)^2 - 1} - (4x+1)\n={(4x+1)(4x-1)} - (4x+1)\n=(4x+1)(4x-2)\n= 2*(4x+1)(2x-1)
1099193
I assume you mean you can only use the symbols and digits once to make it equal two.  In other words, use 1 2 3 4 ( + ) - and make it equal two.\n\nThat's easy!\n (4 + 1²) - 3 = 2\n\nHere are some other ways I can think of:\n4 + (1² - 3) = 2\n(4 - 3) + 1² = 2\n(1² + 4) - 3 = 2\n(-2 + 4) 1³ = 2\n1 + (4 - 3)² = 2\n1 + (3 - 4)² = 2 (since (-1)² = 1)\n2 ( -3^1 + 4) = 2 (imagine the 1 is a superscript 1st power)\n(2 + 4) 3^-1 = 2 (imagine the -1 is a superscript, so 3^-1 = 1/3)\n+2³ (4^-1) = 2 (-1 superscript, so 4^-1 = 1/4)\n2(+3) - 4^1 = 2 (imagine the 1 is a superscript again)\n(3 + 1^4) - 2 = 2 (imagine the 4 is a superscript 4th power)
970698
(a + b)(c + d + e)\n= a(c + d + e) + b(c + d + e)\n= ac + ad + ae + bc + bd + be\n\nOR\n\n(c + d + e)(a + b)\n= c(a + b) + d(a + b) + e(a + b)\n= etc.
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
119559
its (x+2)(3x+1) now thats rite lol
985787
this is\n\nx^4+ 4\n= (x^2)^2 + 2^2\nadd and subtract 4x^2\n= (x^2)+2(x^2)*2 + 2^2 - 4x^2\n= (x^2+2)^2 - (2x)^2\n= (x^2+2x+2)(x^2-2x+2)
534750
1/2 + 1/2b + 2b^2 + 6bc + c^2\n\n(1/4b)(1/4b) + 4bc + (bc)(bc)\n1/2 + 1/4b +1/4b + b^2 +4bc + b^2 + c^2 + 2bc\n1/2 + 1/2 + 2b^2 + 6bc + c^2\n\nI think this is right... email me if I'm wrong?
1023705
2 + 2 / 2 =\n2 + 1 = 3
471544
Technically it is 3\n\nMultiplication/division always comes before addition/subtraction, so that question becomes half of two (1) plus 2 (1+2=3).\n\nIf you were to bracket the (2+2) on the other hand, then the equation would become half of (2+2) = half of 4 = 2\n\n\nEdit: Haha, cheers aphanes.
1027362
(1+2)/2=3/2\n1+(2/2)=2
758567
2(2x-4)= x+4\n4x-8 = x+4\n4x-x = 8+4\n3x = 12\nx=4\n\n&\n\n4(x+3)+3= 5(x-4) + 5\n4x+12+3 = 5x-20+5\n4x+15 = 5x-15\n4x-5x = -15-15\n-x= -30\nx = 30\n\n&\n\nx+2= 4(5x-2)\nx+2 = 20x-8\n2+8=20x-x\n10 = 19x\n19x= 10\nx=10/19\n\n&\n\n21-3x = 4x+7\n21-7 = 4x+3x\n14=7x\n7x = 14\nx=14/7\nx=2