Prove the 2nd part of the Diophantien equation theorem:  That if Xo, Yo is on solution, then every solution is of the from: X=Xo+(b/d)t;  Y=Yo-(a/d)t for t is an integer.\n\nMaybe if you knew that the first part of the equation is: ax+by=d which I already proved.
216451
First do this for d=gcd(a,b)=1. Subtracting the equations for the solutions (x_0,y_0) and (x,y), get that a(x-x_0)+b(y-y_0)=0. If gcd(a,b)=1, this shows that a divides y-y_0, i.e. y=y_0+at. Conclude that x=x_0+bt.\n\nNow see what happens if gcd(a,b)=1 and d is not 1. Then do the general case.
235793
Hmmm... what two numbers can you plug in for x and y to have x/y + y/x = 1, anyway? I have the oddest feeling that no two numbers would fit the initial criteria. That would explain the paradox...\n\nAlternately, you could try working backwards. Suppose x=3 and y=-3.\n\nAs 3³+(-3)³=0, these two numbers work for at least part of the problem. NOW comes the working backwards part.\n\nBecause of the "sum of two cubes" property, you can turn 3³+(-3)³ into [3+(-3)][3²+(-3)²-3×(-3)] (This is the equivalent of the step after "Therefore" on the question. Also, note that × (NOT x) represents the multiplication sign.)\n\nSo, you then have [3+(-3)][3²+(-3)²-3×(-3)]=0. You'd normally divide both sides by xy (which is 3×(-3)), the reverse of what was on the question (as we're still working backwards), and you'd have 3²+(-3)²-[3×(-3)]=0. (Same as x² + y² - xy = 0.) HOWEVER, xy=0. Doing this operation backwards would involve dividing by zero (a big NO-NO in math), and so doing it normally involves multiplying by zero, and weird things can happen when you solve equations by multiplying by zero. For one thing, an untrue equation can turn true. See below:\n\n3≠5, but\n3×0=5×0 because anything times 0 is 0.\n\nSo, if the top equation (x/y + y/x = 1) is untrue, is could've turned true when you multiplied by zero. Now, I can't prove it, but I'm almost certain that there aren't any two numbers you can plug in for x and y in the above equation. So, it's untrue. And so, anything based on it (unless you multiplied it by zero) is also untrue. So, the "But we assumed in the beginning that x/y + y/x = 1" part throws everything off, because x/y + y/x can't equal 1.\n\nSo, there's my explaination. A little lengthy, I know. It ain't pretty, but (hopefully) it gets the job done...
60530
No, nobody really can prove it, with assuming several axioms\nfirst. There are certain things you have to assume are true in your universe, without necessary being able to check them. From those axioms, you can derive conclusions (theorems) that are always true, but only in that particular\nuniverse. The discussed operation of addition is one such theorem. If you assume that the axioms of natural numbers are true, then this addition becomes quite trivival. If you assume those axioms of natural numbers to be false, then the above operation of addition might become false with a different set of supposition. In other words, only with the facts given, one cannot prove, nor disprove, the fact of addition. It all depends on the set of assumptions you would have made before adding the two ones.
533824
Fermat's Last Theorem states that \nx^n + y^n = z^n \n\nhas no non-zero integer solutions for x, y and z when n > 2. \n\nHis proposition was about an equation which is closely related to Pythagoras' equation. Pythagoras' equation gives you:\n\nx^2 + y^2 = z^2\n\nYou can ask, what are the whole number solutions to this equation, and you can see that:\n3^2 + 4^2 = 5^2\nand\n5^2 + 12^2 = 13^2\n\nAnd if you go on looking then you find more and more such solutions. Fermat then considered the cubed version of this equation:\nx^3 + y^3 = z^3\n\nHe raised the question: can you find solutions to the cubed equation? He claimed that there were none. In fact, he claimed that for the general family of equations:\nx^n + y^n = z^n where n is bigger than 2\n\nit is impossible to find a solution. That's Fermat's Last Theorem.\n\n\nProof : \n\nn = 3\n\nx^3 + y^3 = z^3 has integer solutions -> xyz = 0\n\n(1) Let's assume that we have solutions x,y,z to the above equation.\n\n(2) We can assume that x,y,z are coprime. \n\n(3) First, we observe that there must exist p,q such that \n\n(a) gcd(p,q)=1\n(b) p,q have opposite parities (one is odd; one is even)\n(c) p,q are positive.\n(d) 2p*(p2 + 3q2) is a cube.\n\n(4) Second, we know that gcd(2p,p2+3q2) is either 1 or 3. \n\n(5) If gcd(2p,p2+3q2)=1, then there must be a smaller solution to Fermat's Last Theorem n=3. \n\n(6) Likewise, if gcd(2p,p2+3q2)=3, then there must be a smaller solution to Fermat's Last Theorem n=3.\n\n(7) But then there is necessarily a smaller solution and we could use the same argument on this smaller solution to show the existence of an even smaller solution. We have thus shown a condition of infinite descent.\n\n\nI can not have much explaination about in general n. Anyways, that was a real nice question.\n\nI read this somewhere, its not my creation. \n\nAll the best and have nice time
90848
You finally have xy=0 so x=0 or y=0 (unless it is a problem of modular algebra...)\n\n----\n\nNotice that xy doesn't "cancel": you have \n45xy=148xy\n\nBring all terms on the left:\n(45-148)xy = 0 ----> -103xy = 0\n\ndivide by -103:\nxy=0\n\nby the law of cancellation, the last equation is true for\n(x=0 and y whatever) OR (x whatever and y=0)\n\nSo there are infinite solutions for the couple (x,y)\n\n\nGraphically, the solutions are given by all the points of the x and y axes
520435
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
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its simple distributive law.\nwhat has happened is they have taken (a-2) common out of both the terms .\neg: cx-x = x(c-1)\nthe same thing has happened here with x=(a-2) and c=a^2\nso, in step 2, we have: a^2(a-2)-(a-2)\ntaking (a-2) common outside; you end up with (a-2)*((a^2)-1)\nwhihc is [3]\nthen as you know, (a^2)-1 is same as (a^2)-(1^2) and then we apply the algebraic identity (a^2)-(b^2)=(a+b)*(a-b) to get (a-2)(a+1)(a-1) which is [4]
552507
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
66997
Equation #1 >>> x^2+y^2=1\n\nLet the value S=(1+x+iy)/(1+x-iy)\nmultiplying by the conjugate of the denominator (1+x+iy)\nS=[(1+x+iy)(1+x+iy)]/ [(1+x-iy)(1+x+iy)]\nS=[1+x+iy+x+x^2 +ixy+iy+ixy-y^2]/ [(1+x)^2-(iy)^2]\nnote:i^2=-1 and (x+y)(x-y)=x^2-y^2 but (x+iy)(x-iy)=x^2+y^2\n\nS=[1+x^2+2x+2iy+2ixy-y^2]/ [1+2x+x^2+y^2]\nsubstitut in the numerator 1-y^2=x^2 from eqn #1\nsubstitut in the denominator x^2+y^2=1 from eqn #1\n\nS=[2x^2+2x+2iy+2ixy]/[2+2x]\nDivide by 2 on the num.and den.\n\nS=[x(1+x)+iy(1+x)]/[1+x]\ndivide by (1+x) on the num and den.\n\nS=x+iy\n\nthis is the value required
261666
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
464332
The proposition is\n\nSQRT(a) + SQRT(b) = SQRT(a+b)\n\nIf the equality is true, then we can rais both sides to the second power, yielding\n\n(SQRT(a) + SQRT(b))^2 = (a+b)\n\nIf we put c=SQRT(a) and d=SQRT(b) to simplify the equation, we have \n\n(c + d)^2 = a+b\nand expanding\nc^2 + 2cd + d^2 = a+b\n\nBut c^2 = SQRT(a) ^2, and d^2 = SQRT(b) ^2, so in the end\nc^2 = a and d^2 = b\nso back substituting we have\na +2cd +b = a+b\nand 2cd must be equal to zero, and that is possible only if either c or d is zero, and that is possible only if either a or b is zero.
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It is false no solution\n\nLook at \n Example 5:  on this site\n http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut14_solve.htm\n \nWhere did our variable, y, go???  It disappeared on us.  Also note how we ended up with a FALSE statement, 0 is not equal to 11.  This does not mean that y = 11 or y = 0.  \nWhenever your variable drops out AND you end up with a false statement, then after all of your hard work, there is NO SOLUTION. \n\nSo, the answer is no solution.
654398
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
453890
1) (I had had another proof but I like this one better)\nlet x = 4k+a for k = 0,1,2,3,... and a = 0,1,2, or 3\nS = 1^x + 2^x + 3^x + 4^x = 1 + 2^(4k+a) + 3^(4k+a) + 4^(4k+a)\n= 1 + 2^a(2^4)^k + 3^a(3^4)^k + 4^a(4^4)^k \n≡ 1 + 2^a(1)^k + 3^a(1)^k + 4^a(1)^k (mod 5) by Fermat's theorem\n≡ 1 + 2^a + 3^a + 4^a (mod 5)\n= 10, 30, 100 ≡ 0 (mod 5) i.e., S divisible by 5 when a = 1,2,3 \n≡ 1+1+1+1 ≡  4 (mod 5) i.e., S not divisible by 5 when a=0, i.e, x divisible by 4\n \n2)\nx^2 - y^2 = 12^2 = 144\n(x+y)(x-y) = 144\nThe 15 factors of 144 are 1,144; 2,72; 3,48; 4,36; 6,24; 8,18; 9,16; 12\nx+y and x-y will be integers only when cofactors are both even or both odd.\nThis leads to (37+35)(37-35); (20+16)(20-16); (15+9)(15-9);\n(13+5)(13-5)...[i.e.,4 cases (x,y) = (37,35), (20,16), (15,9), (13,5)]\n\n3)\nIf x is even (x=2k), x^4 + 4^x = (2k)^4 + 2^(2x) and is divisible by 2, so not prime.\nIf x is odd (x=2k+1), then\nx^4 ≡ 1 (mod5) for any x by Fermat's theorem,\nand 4^(2k+1) ≡ 4*4^2k ≡ 4*2^4k ≡ 4*(2^4)^k ≡ 4*1^k ≡ 4 (mod 5)\nx^4 + 4^x ≡ 1+4 ≡ 5 ≡ 0 (mod 5) and is thus divisible by 5 , so not prime.\n\nI hope you understand these and are not just handing it in.\nIf there's a next time please ask three questions.
406006
So you need to prove that each element (x,y) E br(xo, yo) has a neighbourhood contained in br(xo, yo),\n\nTake a point (xa,ya) E br(xo,yo) and let d = r - ||(xa,ya) - (xo,yo)||.  d is positive since ||(xa,ya) - (xo,yo)|| < r.\nThen the neighbourhood bd(xa,ya) satisfies the condition because for any point (x,y) in bd(xa,ya) we have\n||(x,y) - (xo,yo)||\n<= ||(x,y) - (xa,ya)|| + ||(xa,ya) - (xo,yo)|| (triangle inequality)\n< d + ||(xa,ya) - (xo,yo)|| (because (x,y) E bd(xa,ya))\n= r.\n\nIn plain English: If the point (xa,ya) has distance d from the boundary of the set, then a neighbouroud of (xa,ya) whose elements are all less than d away from (xa,ya) will be contained in the set as well.
143801
Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
430048
1.) (x^2y)/(xy) = ((x)(x)(y))/(x)(y) = (x/x)(y/y)(x) = (1)(1)(x) = x\n\n2.) (3(x-2)^2)/(6(x-2)^4) = ((3)(x-2)(x-2))/((3)(2)(x-2)(x-2)(x-2)^2)\n    *Now what can be "1" \n\n3.) There may be a type-o in the denom having the "+" sign, but if such the case, the expression is at it's simplest form.\n\n4.) Now for this think about factoring out a number and I think you'll see something that can be in its simplest form.