Number Sense: Can 1+1+1=3 be written as a multiplication sentence of equal groups?
958363
Ummm, I'm not even sure what that means, but I'm guessing:\n\n(1 *1) + (1*1) + (1*1) \nerm...or, in sentence form one times one plus one times one plus one times one.
1130664
1+1 = 10 if you are counting in binary.\n\n1+1=0 with 1 carry over if you are counting in urinary.\n\nanything bigger than binary = 2.\n\n1+1 = 1 if you are talking logical (OR) also.\n\n1+1 = 2 at first then it becomes 3 then 2 again if you are talking about marriage and divorce.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
932028
Yes.\n\nThe argument is valid but not perfectly worded.  \n\nHere A, B, C can be variables as others have pointed out or they are statements that are fulfilled in direct progression.  In fact this is a form of the "Direct Proof"\n\nDirect Proof example:\n\nAs an example we can claim that if we have A:={the addition of two odd numbers x, y |  x+y} it will become an even number which can be statement C.\n\nSo by our knowledge that odd numbers can be written as \nx = 2m+1   and y = 2n+1  where m, n are any integers we can use that as statement B.  Hence B:={x = 2m+1   and y = 2n+1  where m, n are any integers}\n\nDoing a little algebra we have \nx+y = (2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)\n\nNow we can find out that (m+n+1) is an integer as by definition and we can call it "p" s.t. p=(m+n+1),\nHence x+y = 2(m+n+1) = 2p, and by our knowledge of numbers 2p is an even number.  Hence C is fullfilled.\n\nThus like questioner's argument:\nIf A then B then C\nA \ntherefore C.\n\nAnd with my example and statements in quotations can be be replaced by letters A, B, and C: \nIf "you add two odd numbers" then "they are integers that can be written as the addition of two times another integer plus one which algebraically can be written as two times integer" then "you obtain a number which is even".\n"You add two odd numbers" \ntherefore "you obtain a number which is even".\n                                              Q.E.D.
850001
It's pretty obvious if you write out the multiplication:\n\n......1111\n....x1111\n-----------\n......1111\n....1111\n..1111\n1111\n----------\n1234321\n\nAlgebraically 1111 x 1111 = 1111 x 1 + 1111 x 10 + 1111 x 100 + 1111 x 1000 = 1 x 1 + (1+1) x 10 + (1+1+1) x 100 ....\n\nEach row is 1111 and each row shifts one to the left.\n\nThe nice symmetry would break down after 111111111, because then you'd have to start carrying.
1091111
1, 2, 3, 4, 5,  and so on are all just labels for items.  We have defined the group of integers by 1+1=1'=2, 2+1=2'=3, and so on where for any n≠0 n•1=1+1+1+ . . . +1+1≠0 (with n 1s).  Then we have the negative integers so that for every number a, there is a solution to the equation x+a=0.\n\nOne thing that we can do instead (as is learned in slightly more advanced algebra) is basically set all of the integers equal to 0.  Then 0=1=2=3=4=5= . . .  If we do this, and keep the standard addition that we use everyday, we still have a mathematical object called a group.  It is just like the integers, except that the integers are said to be a abelian group of infinite order, and the group where all the integers are set equal is an abelian group of order 1.  In particular, 1=2.  It is perfectly valid, but since the group of order 1 isn't really that interesting.  But to answer your question, yes 1 can be equal to 2, just not in a field of characteristic 0.
523701
If you add the numbers together then they equal an even number.   1 1  1+1=2
751280
multiplication\n\n\nDoug
161644
I'm trying to take them.  1 + 1 = 3.  Man + Woman equals baby.
1117885
23\n\nAny number exactly divisible by three can be represented just using the three. Any number that yields a remainder of 1 when divided by three and is at least 13 can be represented as 13 plus some number of 3s. Any number that yields a remainder of two when divided by three and is at least 26 can be represented as 13+13+some number of threes. Because all numbers must have a remainder of either 0, 1, or 2 when divided by 3, all numbers at least 26 can be represented with one of these schemes. It is easily seen that adding the ability to include more than two 13s, or any number of sixes, adds no additional power to the system. Therefore, the greatest number not representable would be the largest number less than 26 and having a remainder of 2 when divided by three. This number is 23.
119559
its (x+2)(3x+1) now thats rite lol
599324
order of operations is important but it's impossible to know what exactly do you mean when you write words instead of math symbols, so I would guess that you mean this\n1+1/1-1*1/1=\n\n1+1-1=\n\n1+0=\n\n1
351083
In A3, put "=SUM(A1:A2)" or just "=A1+A2"
771335
Is it twice the "same number + 3"\nor\ntwice the "same number", + 3?\n\nCase 1:\n5(x - 1) = 2(x + 3)\n5x - 5 = 2x + 6\n3x = 11\nx = 11/3\n\nCase 2:\n5(x - 1)  = 2x + 3\n5x - 5 = 2x + 3\n3x = 8\nx = 8/3\n\nThe phrasing can be interpreted either way, so the problem is ambiguous, but I would go with number 2, that seems more intuitive to me.
496228
1 + 1 = 2 ... traditional addition\n\n1 + 1 = 10 ... addition, but in binary\n\n1 + 1 = 1 ... boolean algebra\n\n1 + 1 = 11 ... Concatenation; although, it's usually with strings rather than numbers.
164692
Yes, it goes like this:\n\n\nStatr out with any two numbers.....a & b.\n\nThen set them equal to each other:\n\na = b\n\nThen proceed as follows.\n\na = b\n\n==> a^2 = ab\n==> a^2 - b^2 = ab - b^2\n==>(a + b)(a - b) = b(a - b)\n==> a + b = b\n\nNow....on the left hand side, since a = b as was staed in the beginning you say:\n\n(b) + b = b\n==> 2b = b\n==> 2 = 1\n\n<done>\n\nVery slick, and very cool...and very wrong.....lol.\n\nThe 'error' comes in when you divide by a - b. Since a = b, you are essentially dividing by zero, and that is a mathematical no-no.\n\nHiope that clears things up for you.
901198
answer is 3s. any no. multiply by 1 is the no. itself.