C B D F Di It' Cue Based Feeding: It'S H I F T E T! How Infants Eat: Fill & Download for Free

Can someone answer? A company is planning to organize 8 lectures A B C D E F G and H for 3 subjects - Quants DI English.Lectures are spread over three days.Quant is to be covered first in 3 lectures followed by English and then DI in 2 lectures. Lectures A C D have to be different days. Lecture B F have to be on same day. But lecture B can not be clubbed with A or G or D. Lecture G and H should come on the same day. Lecture A is a lecture on Quant and lecture C can not be on last day. It is also known that at least 3 lectures on day 1. One book has given data insufficient answer to below two questions 1. Which combination of lectures are arranged on second day of series2. Which are the lectures for D.I. But I think there is data about 3rd day of DI that it has 2 lectures so one can answer. Experts please help.

If you know about complex numbers, there’s a relatively easy way to see how to construct a field of [math]9[/math] elements from a field of [math]3[/math] elements.The starting point, a field of size [math]3[/math], is simply [math]\{0,1,2\}[/math] with addition and multiplication modulo [math]3[/math]. It is variously denoted [math]\text{GF}(3)[/math], or [math]\Z/3\Z[/math], or [math]\mathbb{F}_3[/math], or [math]\Z_3[/math], although this last notation clashes with the notation for the [math]3[/math]-adic integers. I’ll simply call it [math]F[/math][math][/math].Just like the field [math]\R[/math] of real numbers, the field [math]F[/math][math][/math] does not contain a square root of [math]-1[/math]. Quite simply, in [math]F[/math][math][/math], [math]0^2=0[/math], [math]1^2=1[/math] and [math]2^2=1[/math] as well. No square is [math]-1[/math], which is [math]2[/math] in this field.Therefore, just like we build the complex numbers from the real numbers by introducing a symbol [math]i[/math] with [math]i^2=-1[/math], we can build a field [math]F(i)[/math] by doing the exact same thing. We look at expressions [math]a+bi[/math] with [math]a,b \in [/math][math]F[/math][math][/math] and [math]i^2=-1[/math]. Addition and subtraction are obvious:[math](a+bi)+(c+di) = (a+c) + (b+d)i[/math][math]-(a+bi) = (-a) + (-b)i[/math]Multiplication is also quite easy. Just expand the brackets as usual, and use [math]i^2=-1[/math].[math](a+bi)(c+di) = ac+(ad)i+(bc)i+(bd)i^2 = (ac-bd)+(ad+bc)i[/math]For division, we need a multiplicative inverse for [math]a+bi[/math] whenever it is nonzero. We use the usual trick of multiplying numerator and denominator by the “conjugate”:[math]\frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2+b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i[/math]Remember: [math]F[/math][math][/math] itself is a field, so expressions like [math]\frac{a}{a^2+b^2}[/math] make perfect sense in [math]F[/math][math][/math], provided that [math]a^2+b^2 \ne 0[/math]. Luckily, owing to the fact that [math]F[/math][math][/math] lacks a square root of [math]-1[/math], the sum of squares [math]a^2+b^2[/math] is indeed never zero unless both [math]a[/math] and [math]b[/math][math][/math] are zero. If [math]a[/math] is nonzero then [math]a^2=1[/math], and if [math]b[/math][math][/math] is nonzero then [math]b^2=1[/math], so the sum [math]a^2+b^2[/math]is either [math]1[/math] or [math]2[/math] but never [math]0[/math].So, [math]F(i)[/math] is a field. How many elements does it have? Well, in [math]a+bi[/math], there are [math]3[/math] options for [math]a[/math] and [math]3[/math] for [math]b[/math][math][/math], so [math]9[/math] options overall. The elements are[math]0,1,2,i,1+i,2+i,2i,1+2i,2+2i[/math]You may prefer to think of those as[math]0,1,-1,i,1+i,-1+i,-i,1-i,-1-i[/math]And there’s your field of [math]9[/math] elements.It is a non-obvious fact that there is only one field with [math]9[/math] elements (or any other order), up to isomorphism. You can construct this field in other ways, but you’ll always end up with essentially the same thing.From a slightly more general perspective, what we did here is form the splitting field of [math]X^2+1[/math] over [math]F[/math][math][/math]. In general, to make a finite field with [math]p^n[/math] elements, you can start with [math]F=\text{GF}(p)[/math], find an irreducible polynomial [math]f(X)[/math] of degree [math]n[/math] over [math]F[/math][math][/math], and look at [math]F[X]/\langle f(X) \rangle[/math]. It’s a fact that there are such polynomials available of any degree over any finite field, and it’s also a fact that the resulting field [math]\text{GF}\left(p^n\right)[/math] is unique, up to isomorphism. Succinctly, there’s always exactly one finite field with [math]p^n[/math] elements, for any prime [math]p[/math] and positive integer [math]n[/math]. There are no fields with [math]N[/math] elements when [math]N[/math] is not a prime power.

Only if the coefficients of the polynomial are real. Otherwise consider [math]p(x)=x-i[/math]. [math]i[/math] is definitely a root of this polynomial while [math]-i[/math] is not.First let’s prove that [math]f(z)=z\bar[/math] is a endomorphism (actually an isomorphism, but we do not care) of [math]\[/math][math]C[/math][math][/math][math]\overline{(a+bi)+(c+di)}=\overline{(a+ c)+(b+d)i}=(a+c)-(b+d)i=(a-bi)+(c-di)=\overline{a+bi}+\overline{c+di}[/math][math]\overline{(a+bi)(c+di)}=\overline{(ac-bd)+(ad+bc)i}=(ac-bd)-(ad+bc)i=(a-bi)(c-di)=\overline{a+bi}\cdot\overline{c+di}[/math]Consider [math]p(z)=a_0+a_1z+\dots+a_nz^n[/math]. Conjugate it and get [math]\overline{p(z)}=\bar{a_0}+\bar{a_1}\bar{z}+\dots+\bar{a_n}\bar{z}^n[/math]. I.e. [math]\overline{p(z)}=\bar{p}(\bar{z})[/math].But if the coefficients of your polynomial are real then [math]\bar{p}(z)=p(z)[/math].If [math]w[/math] is a complex root of a polynomial with real coefficients having a non-zero imaginary part then [math]p(w)=0\Rightarrow \overline{p(w)}=0\Rightarrow \bar{p}(\bar{w})=0\Rightarrow p(\bar{w})=0[/math]. Thus [math]\bar{w}[/math] is also a root.Not only every root w has its pair [math]\bar{w}[/math], but they have the same mutiplicity, because [math]p(z)=q(z)(z-w)(z-\bar{w})=q(z)(z^2–2\mathop{\mathrm{Re}} w+|w|^2)[/math]. And thus [math]q(z)[/math] is also a polynomial with real coefficients. So either it has either both w and [math]\bar{w}[/math] as roots or none of them.