Also, a is NOT a square mod p.\nQ = rationals
125374
holy crap........i never even passed algebra 1 \nsorry
366035
False if a= b^2 else\n\nthis is true(with one additionall exeption a= b^2 and n= 1 satisfiles the condition so a should not be b^2)\n\nif (a)^n = b^(n+1)\n\nthen b = (a/b)^n\ndeviding both sides by b^n\nnow let  a and b hace gcd t\n\na= xt\nb = yt\nand x and y are co-primes\n\nyt = (x/y)^n\n\nLHS is integer and RHS is not an integer as x y are coprimes.\n\nso a contradiction unless x = y\nin tnis case yt = 1\nso t and y and x all are 1\nso a=b=1 holds\n\na=b=0 is a trivial case as both are zero\nso the statement is correct when a is not multiple of b\n\nin case a is multiple of b say \n\na= bt\nb = t\nso when a = b^2 \nnow a^n = b^2n\nso a = b^2\n\nfor all other cases it is true
110986
square root of 3 is an irrational number\nirratioal number can't be described as fraction
276179
Part of your question says, "...q*0/0...," which is of course irrational. Your question needs revision or clarification of some kind.
41436
2+2=1 (mod 3)\n\nNothing is flawed.\nIn natural numbers we always have 2+2=4.\n\nPS.\nJust as a joke for mathematical analysis students:\n2+2=5 for sufficiently large values of 2
486304
all the questions relate to the same basic problem. When a function p(x) changes sign in the interval [a,b] it has to cross the real line (y=0). Therefore it has a real solution there. So, first question is solved since p(1) is negative and p(2) is positive.\n\nthe second question has the same answer, only more complicated but just a change of sign when you consider the polynomial function f(x) - g(x).\n\nthird. all non-real solutions come in pairs therefore an odd degree polynomial has at least one real solution since the number of solutions equals the number of roots. (Fundamental theorem of algebra.)
137705
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
253607
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
510559
the fallacy is square root of (-1)^2=-1 and not mod -1
612268
See, first let us see what exactly is meant by an irrational no. I rrational nos. are non-terminating non-repeating numbers such as 1.123456789637631...... Numbers such as 0.33333333..... are rational and not irrational. \n\nRational numbers are those numbers which can be expressed in the form of p/q where q not equal to 0.  0.33333333....  is rational since it can expressed as 1/3 . \n\nAccording to ur ques., i think 0.375932667 is a rational numbr since it can be represented as 375932667/1000000000.\n\nBut if it was 0.375932667.....  that wud become non-terminating non-repeating no. thus it wud be an irrational number.\n\n\nAlso, 0 is a rational as well as a real number.
263579
a)\ni: s may not be zero, as m/s would be undefined\nii: s=m/(x-z)\nx may not equal z, or you'd be dividing by zero again...\niii: x=z+m/s\nas part i, s may not be zero\n\nb) √(x²-4) may not be zero, so x may not be ± 2\nfurther, if -2<x<2, then √(x²-4) will be a complex number, so for both T and A to be real, x would have to be greater than 2 or less than -2...\n\nif the subject is changed to x:\nx=√[(A/T)²+4]\nthen T can't be zero
456262
Hello Rajeash\nYour additional question "\nProve that when x is even then it is impossible to write the above equation For n>2.\n"\nIsnt this what thye good o'l  Sir Fermat already has stated even for odd x ?\n\n\na^(n) - b^(n) = x^(n) \nsuppose that n is not prime, thus n = p*q, with p a prime\nthen we can write\n(a^q)^p -(b^q)^p = (x^q)^p \nso we can limit our problem to n a prime. Fermat stated his statement for p>2 thus we only have to investigate the odd primes.I do not agrede with You Rashesj that we have to investigate 4 as a separate case, because 4 = 2^2 we can write the equation as (a^2)^2 - etc , Do you agree with this ?\n\nYou also say that we can assume that if we have a solution then x will be an odd prime. If a,b,x have a common factor then we can divide that factor out. So we can assumet that they dont have a common factor. If x is even then a and b should be odd.
66911
The construction of negative numbers and rational numbers is much more complex than you think. So 3 over -1 is not an arithmetic operation and using the rules of abstract mathematics it can easily be seen it is true.
490941
A=3*sqrt3 *a^2\nhere a =4\nso A= 3*sqrt3*16\n= 48*sqrt3\nor= 16*2.59808\n= 41.56928 sq.m
178383
Seems like I have been doing most of your homework, SMILE4LISA *wink.  But I do not mind.  I noticed a couple of errors in your question, so I have taken the liberty of fixing them myself.\na) Should read: factor a^6 + 1 = (a^2)^3 + 1 as a sum of two cubes.\nANSWER: Using the formula, a^3 + b^3 = (a + b)(a^2 - ab + b^2), I can write a^6 + 1 = (a^2)^3 + 1^3.  Hence, a^6 + 1 = (a^2)^3 + 1^3 = (a^2 + 1)((a^2)^2 - a^2 + 1)\nb) Simple plug n chug: \n2^6 + 1 = (2^2)^2 + 1^3 = ((2^2) + 1)((2^2)^2 - 2^2 + 1), both check out to be 65 when simplified.\n2^12 + 1 = (4^2)^3 + 1^3 = ((4^2) + 1)((4^2)^2 - 4^2 + 1), both check out to be 4097.\nc)It should read: show that a number of the form 2^n + 1 (n > 0) cannot be prime if n is a multiple of 3 (ie if n = 3k for some integer k)\nANSWER: 2^n + 1 = 2^3k + 1 = (2^k)^3 + 1^3 = (2^k + 1)(2^2k - 2^k + 1) and this implies that it is divisible by the factors of (2^k + 1) and (2^2k - 2^k + 1); primes can only be divisible by 1 and itself.
491463
Necessary and sufficient condition is a is not the neutral element  and m is a prime superior to 2
284671
possible rational factors are p/q where, p is factors of the constant and q is factors of the leading coefficient.\np = 15 and q = 2\nfactors of p: 1, 3, 5, 15\nfactors of q: 1,2\nPossible p/q = (plus or minus) 1, 3, 5, 15, 1/2, 3/2, 5/2, and 15/2\nKeep picking one of these values and test it until you get a remainder of 0.\n1 works\n1 2 3 -11 -9 15\n ___ 2  5  -6 -15\n__2 5 -6 -15 0\nThus,\n2x^4 + 3x^3 - 11x^2 - 9x + 15 \n= (x-1)(2x^3 + 5x^2 - 6x -15)\n2x^3 + 5x^2 - 6x -15 can be factored by grouping.\n(2x^3 + 5x^2) + (- 6x -15)\n= x^2(2x+5) -3(2x+5)\n= (x^2-3)(2x+5)\nAnswer: (x-1)(x^2-3)(2x+5)