Explain why it makes sense to declare both n + 0 and 0 + n  = to n?
467241
Add zero to anything and you get the same number you started with. It does not matter at which number you start with add 0 to it, it will still be zero. For example, you can start with 0 and add any number (n), lets say, you will still get that number n. Thus the notation for the previous example would be [0+n]. Ex. [0+1]=1 and again in the notation that you are asking about [0+n]=n, in the example above i just "substituted" 1, or just let [n] that represents any number for instances equal one. \n\nThe same goes for [n+0]=n, let [n] equal 1. add 1 to zero. What do you get? You should get one. see they (n+0 and 0+n )are the same thing.
845089
Ok, your question makes absolutely no sense.  Why say we can only use 1 through 16, and then say we can only use 1?\nSome ways, assuming I translated your question correctly, would include 16 + 16 +1 +1, and 16 + 15 + 3.... There are almost an infinite number of ways to add to 34 with the numbers 1-16, so please rephrase your question.
314173
(n+1)!=(n+1)(n)(n-1)(n-2)...\n  n!  =(n)(n-1)(n-2)...\n\n(n+1)!/n! = n+1\n\n\n(n+2)!=(n+2)(n+1)(n)(n-1)...\n  n!  =(n)(n-1)(n-2)...\n\n(n+2)!/n! = (n+2)(n+1)
719132
"x = x + 1" is the same as "x++" and "++x", the difference is in "x++" and "++x".\n"x++" performs the increment process after the end of the current process while "++x" performs it before the start of the current process, I know you don't understand anything but consider the following example, I guess you will get it:\n\nif myFunction is a function that accepts integers then:\nx = 5;\nmyFuction(x++);\n5 will be sent to myFunction then it will be incremented to be 6 AFTER myFunction is executed, but if it was:\nx = 5;\nmyFunction(++x);\nthen 6 will be sent to myFunction because "x" was incremented BEFORE the myFunction was called.
252229
Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
725270
When an iterative calculations with predefined conditions are to be done, the need of loop arises.\n\nThe following are the loop constructs...\n\n1. While Loop:\n\nSyntax: \n\nwhile( x < 100 ) {\n     if( x < 0 )\n       break;\n     cout << x << endl;\n     x++;\n   }   \n\n2. for loop\n\nSyntax: \n\nfor( initialization; test-condition; increment ) {\n  statement-list;\n  }\n\n   for( int i = 0; i < 10; i++ ) {\n     cout << "i is " << i << endl;\n   }\n   int j, k;\n   for( j = 0, k = 10;\n        j < k;\n        j++, k-- ) {\n     cout << "j is " << j << " and k is " << k << endl;\n   }\n   for( ; ; ) {\n     // loop forever!\n   }            \n\n3. do \n\nSyntax: \n  do {\n  statement-list;\n  } while( condition );
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
357910
hello!!!!!!!!!! \n1+1 in binary is 10! it's not ten but one zero!!!
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
100604
anything like, n.0... , n.1..., n.2..., n.3..., n.4... will be $n.\n\nanything like, n.5... , n.6..., n.7..., n.8..., n.9... will be $n+1.
145210
They call ZERO an even number, who are they that mention like this?\n\nActually, odd numbers are start with 1, then 3, 5, etc., even numbers are start with 2, then 4, 6, etc.\n\nIf you check at the numbers system, then you can understand what meant by odd number, even number, negative number, natural number, whole number and others. \n\nSo, here the point is, ZERO actually is not a even number (the smaller even number start with 2), then there was no reason to tell you that why can call it an even number.
662225
zero is an integer.\nAn integer is any whole number (positive or negative), including zero.
1119693
I think you might need to elucidate more on that question.\n\nPerhaps you mean how do you store multiple values in a data structure.\n\nPerhaps you mean how do you modify a "final" variable. In other languages this is called a constant. The point of this is that it cannot be changed at runtime.
312174
It seems undefined since for x <= 0 and x != odd number, x^x doesn't work out so well.  If it's x->0+, then it's just 1.
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!
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
579225
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero