Prove that 2=1 by Boolean's logic
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In boolean logic, a rule says: A + A = A (see: http://en.wikipedia.org/wiki/Boolean_algebra if you don't believe)\n\nThus if A equals to 1:\n1 + 1 = 1\n2 = 1\n\nNOTE: Wikipedians uses v and inverted v to represents the use of AND and OR logic, I am used to use + and . (multiplication dots), and I think this is the more correct one than wikipedians, since the equation being done here is more in boolean algebra and not only boolean logic
184777
conditional statement and its inverse
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In (a+b)(a-b)=b(a-b)\n\nyou divided by (a-b)\n\nHowever a-b = 0 so you can't divide by it
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1= B\n2= A\n3= D\n4= D\n5= D\n6= C
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=IF(AND(A=[condition], B=[condition]),X,Y)\n\nAND will return true if both conditions are satisfied.
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it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
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we use binary numbers is boolean algebra\n1's and 0's\n\nthe common operators in boolean algebra are\nand (x)\nor (+)\nnot (/)\n\nthink of it as a series of true/false questions\n1 being true and 0 being false\n\nif we use the AND (x) operator\n1 x 1 = 1   in the AND operator, both the first term AND the second term must be true for the answer to be true\n\nthe truth table for the AND (x) will look like\n0 x 0 = 0 \n0 x 1 = 0\n1 x 0 = 0\n1 x 1 = 1  only when both are true we get a 1\n\nfor the OR (+) operator, as long as one of the terms is true, the answer is true. (one OR the other must be true)\n0 + 0 = 0 \n0 + 1 = 1\n1 + 0 = 1 \n1 + 1 = 1\n\nthe not negates the value. /1 = 0 & /0 = 1
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I hope you are trying to say x+2 = (2/3)(2x-1)..if it is correct then\n\n3(x+2) = 2(2x-1) --> 3x+6 = 4x - 2 --> 6+2 = 4x-3x --> 8 = x
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y = |x|   :   x = 0\n\ny = |x| +2   : x = 0\nor\ny = |x + 2|  :  x = -2\n\ny = |x - 3| + 2   :   x = 3
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n, n+1 for every n>=2\n\nif 2 <= 3 then either 2 = 3 or 2^2 = 4 <=3, both of which are false.\nif 3 <= 2 then either 3 = 2 or 3^2 = 9 <=2, both of which are false.\n...\n\nif n <= n+1 then either n = n+1 or n^2 <= n+1, both of which are false for all n>=2.\nif n+1 <= n then either n+1 = n or (n+1)^2 <= n, both of which are false for all n>=1.\n\nThus, n and n+1 are incompartible for all n>=2.
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Try http://www.asic-world.com/digital/boolean.html for a beginning.  I would start with basic boolean math.  That often will help you simplify a large logic circuit down to something you can build.
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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).
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i know how it's done. in the theorm, every number is one higher than stated (ex. 1=2, 2=3, 3=4, etc) since it's 2+2, it's actually 3+3. and if 3+3 =6, and 6=7, then the resulting equation is 2+2=7
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(0! + 0! + 0! + 0!)!\n=(1+1+1+1)!\n=(4)!\n=24
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You prove that for any tiny little e>0 you pick, there exists a little d>0 such that \n\nif abs(1/(x^2)-1/x^2))<e, then abs(x-x)<d,\n\nand since x-x=0 (this is a trivial limit unless you've messed up the parentheses in the problem), just let d=e, or e/2, or any value between e and 0.
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Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
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x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y