Real numbers! Please help! Let  x and y be any real numbers. Prove that exactly one of the following is true:?
148466
You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,
904022
Surprisingly, yes.  This bugged the heck out of me in middle school, but it's a consequence of the way infinity works.\n\nThe classic proof is:\n\n    x = 0.999999....\n10x = 9.999999....\n\nSubtract the first line from the second, and you get\n\n9x = 9, therefore x = 1.
388398
i think the false statement is C. as the degree of any number is always 1. i dont understand the statement b tho :S
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
835878
There's a better way of looking at it.\n\nFor every real number x in [0,1) you define D(x, n) as the nth decimal of x.  This is defined for every natural number n.\n\nFor x=1/3, we have D(1/3, n) = 3 for every n.  The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3.  It is EXACTLY the number 1/3.  Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...\n\nFor 0, we have D(0,n) = 0 for every n.  There is never a '1' decimal.  If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p.  I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1.  That number is greater than zero and smaller than the number that you gave me.  \n\nThe 0.5 = 0.49999... thing is just a quirk of decimal notation.  We can formally prove the equality as follows:\n\nLet a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.\nLet b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.\n\nProof by contradiction that a=b: Let's suppose on the contrary that a > b.  This means there must be a number x between a and b.\n\nLet's figure out the digits of x.  Since x is less that a=0.5, its first decimal must be a 4.  D(x,1) = 4.\n\nThe second decimal of x must be 9 since x > b `> 0.49.\nThe third decimal of x must be 9 since x > b >0.499.\nAnd so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.\n\nTherefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b.  Contradiction, ergo a = b.\n\nThat's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0.  In the latter case, it is possible to find a number that is between the two.
500255
This is called an inequality. First, find the zeros (in this case, -4 and 9). Next, make a number line:\n<------(-4)----------(9)----->\nNow, determine where inside and outside the zeros the inequality will be true or false. Finally, express your answer in terms of the inequality.\nExample: x+12 > 0\nAnswer: x > -12
168972
Your question does not make sense. x cannot be less than or equal to 0 and less than or equal to 7 unless it is less than or equal to 0, which I do not think you mean. \n\nI think you mean 0<=x<=7\n\nThe rest of the question makes no sense. \n\nAnd it really smells like a homework problem. Try thinking for a change.
621381
b) Replace y with 0\n(100+x)^0 = 2*100^0\n1 = 2*1\n1 = 2\nThis implies that y cannot be zero.\nc) Replace x with 0.\n(100+0)^y = 2*100^y\n100^y = 2*100^y\n0 = 2*100^y - 100^y\n0 = 100^y\nThis is never true.\nThis implies that x cannot be zero.\na) The graph looks similar to the graph of y = 1/x
104840
Actually, as what I have learned, this equation is limitless because the numbers itself are limitless. The incredible fact about this is that in every number you substitute in "x", the answer is always "0".
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
939642
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.
163896
yes the previous equation mentioned would "prove" it, but the problem occurs as you're dividing by zero.  you can also make 1=2, with a similar process, involving the division of zero.  if you don't get what i mean, the previous explanation has this:\n\n(x-y)(x+y)=y(x-y)\nx+y=y\n\nwhat happens at this step?  you are dividing by (x-y).  since x=y is given, then (x-y)=0, thus you are dividing by 0, which is impossible in math.
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
759729
I think, therefore I am.\n\nAssume 1 does not exist.\n\nLet x = 0\nLet y = 2\n\nAny pair of numbers have a number between them, known as the arithmetic mean.\n\nTake the arithmetic mean of x and y:\n(x + y)/2 = (0 + 2)/2 = 1\n\nBut we now have a result of a number that shouldn't exist.\n\nSince we have a contradiction, this proves that our initial assumption was incorrect.\n\nTherefore 1 exists -- proof by contradiction.
576554
Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
795059
You came up with x > -1 OR x < 7.  The "or" is key.  It can be less than 7, or  greater than -1.  There's no real number that doesn't fit the solution.\n\nFor the second one, you came up with x >= 0.8 AND x >= 2.  Similarly, the "and" is the key here.  Anything greater than or equal to 2 is going to be greater than 0.8, fulfilling both conditions.  The same cannot be said if the numbers are reversed.\n\nTo sum it up, if the problem has an "or" condition, every solution must solve at least one inequality.  If the condition is "and", every solution must solve ALL inequalities.
955603
x+0=x is always true. This is the definition of the identity property of addition.