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## Four dimensional bailiwick over sophisticated numbers

**This shouldn’t be an retort however an retort to some feedback which is just too lengthy to breathe a statement itself**:

Intuitively this development feels affection a 4 dimensional house, but it surely’s not. Is there some lifelike or layman’s clarification that may make it extra limpid what is definitely taking place on this development?

I assassinate not maintain a graphical clarification. The definitive level certainly is that the house shouldn’t be 4 dimensional. If you might be immediate with group shows, a really related loom happens. Suppose you maintain a gaggle generated by a bunch of components ($a,b,c$) and you might be giving a bunch of relations in between these components ($a^2b=cb, dots$). It is workable that you just maintain so many relations that one does probably not necessity all turbines (for example $a^2b=cb$ implies that $a^2=c$ and thus $c$ is redundant). In this you at first power breathe tempted to arbitrator that this group wanted three turbines, however truly much less suffice.

Your fb man did one thing related. At first graze it seems affection you necessity 4 unbiased vectors te categorical all the things, however these vectors are actually not unbiased. That’s the crux of the rife story.

Or alternatively, what would breathe the implications if we have been to suppose that $z^2=i$ had greater than two options?

If $z^2-i=0$ has greater than two options, you can’t breathe working in a **bailiwick**. Indeed, you’ll be able to simply show that any polynomial $P(X)in Ok[X]$ of diploma $n$ over a bailiwick $Ok$ can not maintain greater than $n$ roots. The proof is easy as quickly as you maintain the division algorithm for polynomials over a **bailiwick**.

**Edit:** I’ll comprise a proof of the latter assertion. Let $P(X)in Ok[X]$ breathe a polynomial. We can write $$P(X)=sum_{i=0}^na_iX^i$$ with every $a_iin Ok$ and $a_nneq 0$. First level to that we could occupy that $a_n=1$ by multiplying the polynomial by $a_n^{-1}$ (right here I make use of that $Ok$ is a bailiwick). Furthermore, doing so doesn’t change the roots of $P(X)$ (that is a pretense that you may show as an relate).

Now I pretense the next: If $cin Ok$ is a root of $P(X)$ (i.e. $P(c)=0$), then $$P(X)=(X-c)Q(X)$$ for some polynomial $Q(X)in Ok[X]$ with $deg(Q(X))=deg(P(X))-1$. Indeed, by the division algorithm for polynomials, one can write $$P(X)=(X-c)Q(X)+R(X)$$ the place $Q(X),R(X)in Ok[X]$ and $deg(Q(X))leq deg(P(X))$ and $deg(R(X))<deg(P(X))=1$. By assumption, $P(c)=(c-c)Q(c)+R(c)=R(c)=0$. On the opposite hand, $deg(R(X))=0$ and thus $R(X)=0$ as it’s a ceaseless polynomial. Thus $P(X)=(X-c)Q(X)$ as required.

Hence we now confirmed that if $c$ is a root of $P(X)$, then we discover that $P(X)=(X-c)Q(X)$ with $deg(Q(X))=deg(P(X))-1$. Now occupy that $P(X)$ has extra roots than its diploma. Iterating the above process, you discover that $P(X)=(X-c_1)(X-c_2)dots (X-c_m)Q(X)$ with $m>deg(P(X))$, however then $deg(P(X))geq m$, a contradiction!

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