Let A denotes the Set of Algebraic Numbers and T the Set of Trancedental Numbers. Discuss the following: •Are A and T closed with respect to addition and multiplication ? •Are A−{0) and T closed with respect to division?


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Question Number 4027 by Rasheed Soomro last updated on 27/Dec/15

$L e t A d e n o t e s t h e S e t o f A l g e b r a i c N u m b e r s$ $a n d T t h e S e t o f T r a n c e d e n t a l N u m b e r s .$ $D i s c u s s t h e f o l l o w i n g :$ $∙ A r e A a n d T c l o s e d w i t h r e s p e c t t o$ $a d d i t i o n a n d m u l t i p l i c a t i o n ?$ $∙ A r e A - {0) a n d T c l o s e d w i t h r e s p e c t t o$ $d i v i s i o n ?$
Commented by Yozzii last updated on 27/Dec/15
$Algebraic numbers are solutions to finite,non−zero polynomials whose coefficients are rational numbers and are in one variable. Thus, x∈A⇒ x∈Q or x∈C−{transendental complex numbers} x∈T⇒x∉A. Note that ∃x∈A which are irrational e.g the solution to x^2 −2=0. Also x∈A could be a trigonometric function applied to a rational multiple of π. e.g cos3r=4cos^3 r−3cosr. let p=cosr and cos3r=0⇒3r=2nπ±(π/2) (n∈Z) r=(1/3)(2nπ±(π/2)) ∴ 4p^3 −3p=0 has solutions p=cos((1/3)(2nπ±(π/2))) (n=0,1) Under the binary operation ∗ for addition, closure of A is provided since A⊂C exhibits closure. T is not closed under ∗ since, for example, −π,π∈T⇒(π)+(−π)=0∉T. Define ⊕ as the binary operation of multiplication. A does exhibit closure by virtue of C exhbiting closure under ⊕ and A⊂C. T is not closed under multiplication since, for 2π,(1/π)∈T, 2π⊕(1/π)=2∉T.$
$A l g e b r a i c n u m b e r s a r e s o l u t i o n s$ $t o f i n i t e, n o n - z e r o p o l y n o m i a l s w h o s e$ $c o e f f i c i e n t s a r e r a t i o n a l n u m b e r s$ $a n d a r e i n o n e v a r i a b l e .$ $T h u s, x \in A \Rightarrow x \in Q o r x \in C - {t r a n s e n d e n t a l c o m p l e x n u m b e r s}$ $x \in T \Rightarrow x \notin A .$ $N o t e t h a t \exists x \in A w h i c h a r e i r r a t i o n a l$ $e . g t h e s o l u t i o n t o x^{2} - 2 = 0 .$ $A l s o x \in A c o u l d b e a t r i g o n o m e t r i c$ $f u n c t i o n a p p l i e d t o a r a t i o n a l m u l t i p l e$ $o f π .$ $e . g c o s 3 r = 4 {c o s}^{3} r - 3 c o s r .$ $l e t p = c o s r a n d c o s 3 r = 0 \Rightarrow 3 r = 2 n π \pm \frac{π}{2} (n \in Z)$ $r = \frac{1}{3} (2 n π \pm \frac{π}{2})$ $∴ 4 p^{3} - 3 p = 0 h a s s o l u t i o n s$ $p = c o s (\frac{1}{3} (2 n π \pm \frac{π}{2})) (n = 0, 1)$ $U n d e r t h e b i n a r y o p e r a t i o n * f o r$ $a d d i t i o n, c l o s u r e o f A i s p r o v i d e d$ $s i n c e A \subset C e x h i b i t s c l o s u r e .$ $T i s n o t c l o s e d u n d e r * s i n c e, f o r$ $e x a m p l e, - π, π \in T \Rightarrow (π) + (- π) = 0 \notin T .$ $D e f i n e \oplus a s t h e b i n a r y o p e r a t i o n o f$ $m u l t i p l i c a t i o n . A d o e s e x h i b i t c l o s u r e$ $b y v i r t u e o f C e x h b i t i n g c l o s u r e$ $u n d e r \oplus a n d A \subset C . T i s n o t c l o s e d$ $u n d e r m u l t i p l i c a t i o n s i n c e, f o r 2 π, \frac{1}{π} \in T,$ $2 π \oplus \frac{1}{π} = 2 \notin T .$
Commented by Rasheed Soomro last updated on 27/Dec/15

${EX}_{PLANATION}^{CELLENT}!!!$
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