Consider the following relation on Z+={1,2,3,...]: a <= b if a=b or a^2 <= b.  What are several pairs of incomparable elements for the relation?
138177
n, n+1 for every n>=2\n\nif 2 <= 3 then either 2 = 3 or 2^2 = 4 <=3, both of which are false.\nif 3 <= 2 then either 3 = 2 or 3^2 = 9 <=2, both of which are false.\n...\n\nif n <= n+1 then either n = n+1 or n^2 <= n+1, both of which are false for all n>=2.\nif n+1 <= n then either n+1 = n or (n+1)^2 <= n, both of which are false for all n>=1.\n\nThus, n and n+1 are incompartible for all n>=2.
553287
(x+y+z)(x+y+z) = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1\n\n(1-x^2 - y^2 - z^2) = 2xy + 2xz + 2yz\n\n(1 - x^2 - y^2 - z^2) \n------------------------ = xy + xz + yz\n.............2\n\nClearly the left hand side is less than 1/2
545212
For each pair (A, A') of elements of B such that A is a subset of A', pick, if possible, an element C is B' so that A is a subset of B is  a subset of A'. If it isn't possible to find a C, don't pick anything. Since there are countable many pairs, (A,A'), there are only countable many C chosen. Let B'' be the collection of these C. Then elements of B'' are open with property P.\n\nTo show that B'' is a base, suppose that a point x is in an open set U. Choose A' in B with p in A' a subset of U. Then choose C' in B' with x in C' a subset of A'. Finally pick A in B with x in A a subset of C'. Now, the pair (A,A') has elements of B' of the appropriate form, so the C associated with (A, A'), satisfies\nx in C a subset of U. Notice that C and C' may be different!
551961
Note: the symbol * means multiplication\n\n1/(3z+5)(2z-7)\n============\n(3z+5)(2z-7) is a short notation for (3*z+5)*(2*z-7)\n\nThis is exactly equivalent to (3*z)*(2*z-7) +(5)*(2*z-7)\nThis is exactly equivalent to 3*z*2*z +3*z*(-7) +5*2*z +5*(-7)\n\nnow lets reorganize all that:\n 3*2*z*z -3*7*z +5*2*z -5*7\nso in the end we get\n 6*z*z - 21*z +10*z -35\nwe then sum the z\n 6*z*z -11*z-35\n\nor in a more compact notation: 6z^2-11z-35\n\n2/ (2-y)(4-y)\n==========\nThis means (2-y)*(4-y)\nyou do like we did in 1/ and you get  y^2-6y+8
371459
The field properties are the properties of real numbers.\n\n(5)(1/5) = 1 is definitely the inverse property of multiplication.\nIf a = b and b = 7, then a = 7 is definitely the transitive property of equality.\n\nThe others are not as definite, and it depends a bit on how your book/teacher interprets and names things.\n\nMost likely the first is the distributive property, because it's equivalent to 6x + 9x = (6+9)x.  (However, substitution is also being used here.)\n\n(1/7)(1/5) = (1/35) is most likely substitution (or replacement, which is a different name for the same property).  That's not really a standard "field property", but it is normally learned together with them.
510108
slider is correct.  \n\nRemember the distributive property?\na(b+c) = ab + ac\n\nThis is using it in reverse.  \n\nIn part A, notice that there is an a in each term, so you can factor out the a's.  \n3a² + 2a = (3a)*a + 2 * a = a(3a + 2).\n\nIn part B, there aren't any letters in common but you can factor out numbers too.  \n12d²+6 = 6*2d² + 6*1 = 6(2d²+1)\n\nThe other parts use similar reasoning.
539533
z=w+whx\nz=w(1+hx)\nz/(1+hx) = w &#92;
522487
You can start from the following identity for set differences which holds true for any sets X, A, and B\n\nX − (A ∪ B)  =  (X − A) ∩ (X − B)\n\nNow, if B is a subset of A, then A ∪ B = A\n\nthe identity above becomes\n\nX − A =  (X − A) ∩ (X − B)\n\nFor this to hold true, X − B must be a superset of (X − A), or in other words, (X − A) must be a subset of X − B
147645
Since [L : K] = [L : E][E : K] and [L : K] is prime, \neither [L : E] = 1 or [E : K] = 1.\n\nThus either E = L or E = K
165661
1st list of elements=\n\n{X u Y}n Z\n{2,4,6,8,10} n {3,6,10,12}\n= {6,10}\n\n2nd list of elements=\n\n{Y n Z}' /X\n{6,10} / X\n{6,10} / {2,4,6,8}\n={10}
362887
A plus B equals C. That's the only algebraic phrase I remember.
163237
Let's start with a definition:\n\nFor a positive integer p, b^p = b x b x ... x b (product of p b's)\n\nFrom this it is easy to simplify b^c x b^d for c and d positive integers because we have the following:\n\nb^c x b^d = (b x b x ... x b) x (b x b x ... x b)\n= product of c b's x product of d b's\n= product of (c + d) b's\n= b^(c + d)\n\nNow what about when we have positive integers, c > d and we want b^c/b^d for b not equal 0.  Well then we have the product of c b's divided by the product of d b's.  Each of the b's in the denominator cancel with on in the numerator so we are left with the product of c-d b's.  \n\nTherefore b^c/b^d = b^(c - d) for all positive integers c > d.\n\nNow we want these general rules to hold for all integers so when c = d, we define b^c/b^d = b^(c-d) = b^0.  But when c = d, b^c/b^d = 1 so b^0 = 1 for all nonzero b.  Similarly, when we have c < d, we want b^(c-d) = b^c/b^d so that since b^c/b^d = 1/b^(d-c) = 1/b^{-(c-d)}, we define b^n = 1/b^(-n) for all negative integers n (and hence easily verifiable for all positive integers as well).\n\nTherefore, so far we have the following for all integers m,n:\n\nb^m x b^n = b^(m+n) (b not equal 0 if m + n < 0)\nb^m/b^n = b^(m-n) (b not equal 0 if m < n)\nb^0 = 1  (b not equal 0)\nb^(-c) = 1/b^c (b not equal 0)\n\nWe want this to generalize to nonintegers as well so we define b^(1/2) for all positive b for instance to be sqrt(b) as b^(1/2)b^(1/2) = b^(1/2+1/2) = b^1 = b and for all positive b, sqrt(b) is defined to be a positive number that when multiplied by iteslf results in b.\n\nLastly, we get (b^m)^n = b^(m x n) by a similar argument.  product of n products of m b's is the same as the product of m x n b's.
435687
You have a quadratic equation whose factors are (z-2) and (z-6); that is, (z-2) * (z-6) = z^2-8*z+12.\nHow does one get this answer? \nWell you have the z^2 term, which tells you there are two factors that we'll call z+a and z+b; now we have to find a and b. What do we know about a and b? We know that the product of z+a and z+b is z^2 + a*z + b*z + a*b, and this means that a*b=12 and a+b=-8. It's easy to see that a=-2 and b=-6 satisfies these equations, but there is a method that works for all cases (even when the answer is not easy to see) called "completing the square" (see ref.). Read the first example at that web page, and I'll list below the equations to substitute for each equation in the example:\nz^2-8*z+12=0 \nz^2-8*z=-12    (eq. 2)\nz^2+2*b*z+b^2=(z+b)^2 (binomial formula)\nz^2-8*z+16=(z-4)^2 (binomial, completed square where b=-4)\nz^2-8*z+16=-12+16=4 (adding 16 to both sides of eq. 2)\n(z-4)^2=4 (from above 2 eqs)\nz-4=+2 and z-4=-2\nz=6 and z=2\nz^2-8z+12=(z-6)*(z-2)\nI hope this helps.
649355
1. 2\n\n2. 2a^2b\n\n3. 6xyz^2(2z-3y)\n\n4. 3a(4a-b+3ab^2)\n\n5.(d+2)(d+3)\n\n6.(x-9)(x-2)\n\n7. (4x-1)(x+3)\n\n8. (4b-9)(4b+9)\n\n9.(y+7)^2\n\n10. (2m+7n)^2
561011
Assuming A, B, and C to be constants, this is an equation that defines a linear relation between variables x and y. There are a number of places where this would appear, but it would be easier to see it in the form y = -(A/B)x + (C/B), its equivalent.
3086
(([(4+4+4).(4+4)]+4)/4)*(4+4)\n\n(4+(4*4)).([(4+4+4).(4+4)]+4)
18562
D. If a variable some times equal to its both of the bounds, it will be mentioned or written as [a,b] generally called as values between a and b inclusive of a and b both, as per the case mentioned here, the variable can take any value between a and b inclusive of a, so a<= variable < b written as [a,b[