Can Simple Algebra  prove 2 = 1. ?
395828
haha, but this does not work. you cannot mathematically divide by (a-b) because this quantity is equal to 0. (1-1)=0. Dividing by zero yields an indeterminate answer.
723313
Yes\n\n(1-2)^2 = (3-2)^2\n\n3 is not equal to one\n\na and c do not have to be equal since the terms are squared. On removing the squares, you have to account for both +ve and -ve roots.
352513
true! Its fairly obvious by definition:\n\nfor example:\n2*5 = 5 + 5 = 2+2+2+2+2\n\nAn easier way to see this is through an invention by John Napier called Napier's Bones. He converted all complicated multiplications into simple addition
360700
a = 1, b = 2, n = 3:\n\n1^2 = 1 = 4 = 2^2 (mod 3) but clearly 1 != 2 (mod 3)
910422
no \n3x-1>1 \n3x>2 \nx>2/3
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
541116
3=2 did get a correct answer...maybe you need to review your rules of square roots.
732898
ab=a+b\nab-a-b+1=1\n(a-1)(b-1)=1\na=(1/(b-1))+1\nIf a is not 1 and b is not 1, then you can find a if you know b and vice versa. Therefore, if a is anything other than 1, you can find b that works and vice versa.
637914
1=1. Simple yet important.
674498
1+1=2. That's it. The answer for everything else would be simply 42
895454
There is no ordered pair {a,b} which solves both equations\n\nif a+b = a-b , then 2b = 0 or b = 0 and a is anything\n\nBut the second equation is not defined for b=0, no matter what a is (specifically, a/b is not defined for b=0)
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
896553
4x^2+8x-x-2\n(4x-1)(x+2)
678157
2x^2(1-2x)=0\n\nx=0\nx= (1/2)
560506
that works... so does this:\n\n-2 x -1 x 1 x 2 x 3 = 12\n\n-2*-1=2\n2*1=2\n2*2=4\n4*3=12
287019
1 + 1 = 2\none plus one = oneone
557831
Yes. You are right, but if and only if the equation MUST be valid for all a & b. The reason is:\n(x+a+b)^2 - x^2 = (x+a+b - x)*(x+a+b + x) = (a+b)*(2x+a+b)\nand\n(x+a)^2 - (x+b)^2 =[x+a -(x+b)]*[x+a + (x+b)]= (a-b)*(2x+a+b)\ncomparing the above, and noting that a+b could not be equal to a-b for all a & b, then 2x+a+b=0 so x=-(a+b)/2\nSimilarly the second equation could be solved by giving square root of both sides:\nx+a = +/-(x+b)\nit is obvious that "+" sign could not be selected since it required that a=b which is against our assumption. So:\nx+a=-(x+b) ==> x=-(a+b)/2