Can anybody know how to prove 0=1?
148878
1power0 = 1\n1power1 = 1\nso that 0=1
469613
two hints:\n\n1) anything raised to the 0 power gives 1\n\n2) anything raised to a negative power means 1/the same thing raised to the positive opposite power.\n\nEnjoy your homework.
1142259
proof :\n\n1 -1 = 1 - 1\nlhs: 1 - 1 \nrhs: 1 -1 \n\nthaking 1 common  from lhs v hav \n1 (1 - 1 )\n\nexpanding rhs by the formula a^2 - b^2 = (a+b)(a-b)\n\n(1-1)(1+1)\n\nthus we hav \n\n1 (1 - 1 ) = ( 1 - 1 ) (1 + 1 )\ndividing througout by ( 1 - 1 )\n\nwe get 1 = 1 + 1\n\n\n\nDoes this help u??
338816
As in many other pseudo-demonstrations of this kind, the error is to divide by zero (2-2) both sides of the identity. You can't do this because the division by zero is an undefined operation.\n\nFrom a well-known logic theorem, starting from a false statement, you can consistently derive any other false statement. This is the common source of many apparent paradoxes.
204914
I only need four 0.\nbinary:0!+0!+0!+0!=100
647936
The answer is 1 - you're raising a quantity to a power of 0, which always results in 1. The quantity within the parentheses contains no division by zero, so I know it's a real number, too.
252734
This is how I remember it...hope it helps.\n\nA Proof\nLet a and b be any two real numbers. Consider the number x defined by\n\nx = ab + (-a)(b) + (-a)(-b).\n\nWe can write\nx = ab + (-a)[ (b) + (-b) ]       (factor out -a)\n  = ab + (-a)(0)\n  = ab + 0\n  = ab.\n\nAlso,\nx = [ a + (-a) ]b + (-a)(-b)      (factor out b)\n  = 0 * b + (-a)(-b)\n  = 0 + (-a)(-b)\n  = (-a)(-b).\n\nSo we have\n      x = ab\nand\n      x = (-a)(-b)\n\nHence, by the transitivity of equality, we have\n\n      ab = (-a)(-b).
882621
My old math teacher showed me this.\n\nSince A=B, then A-B=0, and you can't divide by 0
357910
hello!!!!!!!!!! \n1+1 in binary is 10! it's not ten but one zero!!!
604013
there is no answer becuase it is impossible to divide by 0
266936
Counterexample: 1\n\nJay's proof is false because x(x-1) = 0 just says that x could be either 1 or 0 to satisfy that equation. x was 1 in that case. It does not mean that 1=0.\n\nJZ-LP's proof is false because you can't say let 1=0.\n\nBianca is just plain wrong.
455928
Use Descarte's rule of signs:\n\nf(x) = x^5 + x^4 - x^2 - x + 1\n  To determine negative zeros, plug in a negative and solve.\nf(-) = (-)   (+)   (-)   (+)   (+)\n  Count the number of sign changes.\nThere are 3 sign changes, so you could have 3 or 1 negative real zeros. (The reason you could have 1, is that there might be a negative imaginary zero). You could graph it to see whether you have 3 or 1.
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
678157
2x^2(1-2x)=0\n\nx=0\nx= (1/2)
164065
The trick with "proofs" like these are generally an obscured division by zero. One of the lines seems to work, though when you divide to cancel out some factors, the quantity you're dividing by could/does equal zero. It's a trick and doesn't say anything substantial about mathematics.
1137788
Nope, 0/0 could be 1 but it could also be anything else.   Division is the opposite of multiplication and 0 times anything at all, even infinity, is still zero.
39584
If you derive a constant, you always get 0