Is-it-possible-such-that-two-trancendental-numbers-add-or-multiply-to-give-a-whole-number-

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Question Number 6491 by Temp last updated on 29/Jun/16

Is it possible such that two trancendental numbers add or multiply to give a whole number?

Is it possible such that two trancendental

numbers add or multiply to give a whole number ?

Commented by prakash jain last updated on 29/Jun/16

π is transcendental. (1/π) is also transcendental. π×(1/π)=1

π is transcendental .

\frac{1}{π} is also transcendental .

π \times \frac{1}{π} = 1

Commented by Temp last updated on 29/Jun/16

what if in form ab, b≠a^(−1) ?

w h a t i f i n f o r m a b, b \neq a^{- 1} ?

Commented by Rasheed Soomro last updated on 29/Jun/16

π×(2/π)=2, where (2/π)≠π^(-1)

π \times \frac{2}{π} = 2, w h e r e \frac{2}{π} \neq π^{- 1}

Commented by Temp last updated on 29/Jun/16

The way I see it, the multiplicitive form ab=c, a,b=transendental, c=integer it seems that ab=c only holds true for if a and b share a relationship of: a=(1/b) ∴ab=(b/b)=c I am unsure if for some two different transendental numbers other than when aa^(−1) =1 exist. e.g. eπ=z, z∉Z

The way I see it, the multiplicitive

form a b = c, a, b = transendental, c = i n t e g e r

it seems that a b = c only holds true for if

a and b share a relationship of :

a = \frac{1}{b}

∴ a b = \frac{b}{b} = c

I am unsure if for some two different

transendental numbers o t h e r t h a n when {a a}^{- 1} = 1

exist .

e . g . e π = z, z \notin Z

Commented by Temp last updated on 29/Jun/16

nπ=k, k∈Z, ∴n≠1, n can be trancendental ∴π=(k/n) = false ∴Impossible for different trancendentals to equal an integer excluding cases like π×(1/π). (i.e. k>1)

n π = k, k \in Z, ∴ n \neq 1, n can be trancendental

∴ π = \frac{k}{n} = f a l s e

∴ Impossible for different trancendentals

to equal an integer excluding cases like

π \times \frac{1}{π} . (i . e . k > 1)

Commented by Rasheed Soomro last updated on 30/Jun/16

Say π,e and other transidental numbers, which can not be derived from one another by +,−,×,÷,power,root are said to be root-transedentals. π,−π,π^(−1) ,nπ,((mπ)/n)(m,n∈Z) are of same roots.e and π are root-transcidentals. Now what you want to say, can be said “Sum or product of two transidentals of different roots can′t be whole numbers” Am I right?

S a y π, e a n d o t h e r t r a n s i d e n t a l n u m b e r s,

w h i c h c a n n o t b e d e r i v e d f r o m o n e a n o t h e r

b y +, -, \times, \div, p o w e r, r o o t a r e s a i d t o b e r o o t - t r a n s e d e n t a l s .

π, - π, π^{- 1}, n π, \frac{m π}{n} (m, n \in Z) a r e

o f s a m e r o o t s . e a n d π a r e r o o t - t r a n s c i d e n t a l s .

N o w w h a t y o u w a n t t o s a y, c a n b e s a i d

“ S u m o r p r o d u c t o f t w o t r a n s i d e n t a l s

o f d i f f e r e n t r o o t s {c a n}^{'} t b e w h o l e {n u m b e r s}^{″}

A m I r i g h t ?

Commented by prakash jain last updated on 29/Jun/16

ab=n⇒b=(n/a) So if product of any 2 numbers is a whole number then second number b=a/n.

a b = n \Rightarrow b = \frac{n}{a}

So if product of any 2 numbers is a whole

number then second number b = a / n .

Answered by malwaan last updated on 02/Jul/16

e^(π(√(163))) is an integer e^(π(√(163))) = 262 537 412 640 768 744

e^{π \sqrt{163}} i s a n i n t e g e r

e^{π \sqrt{163}} = 262 537 412 640 768 744

Commented by malwaan last updated on 03/Jul/16

According to WolframAlpha e^(π(√(163))) =262537412640768743.99999999999925007.... what do you think? very small error because of calculation rounding

A c c o r d i n g t o W o l f r a m A l p h a

e^{π \sqrt{163}} = 262537412640768743.99999999999925007 \dots .

w h a t d o y o u t h i n k ?

v e r y s m a l l e r r o r b e c a u s e o f c a l c u l a t i o n r o u n d i n g

Commented by prakash jain last updated on 02/Jul/16

According to Wolfram alpha e^(π(√(163))) is a transcendental number. 2.6253..×10^(17)

According to Wolfram alpha e^{π \sqrt{163}} is

a transcendental number .

2.6253 . . \times 10^{17}

Commented by Rasheed Soomro last updated on 03/Jul/16

According to wikiepedia e^(π(√n)) is transcedental for any positive n

A c c o r d i n g t o w i k i e p e d i a

e^{π \sqrt{n}} i s t r a n s c e d e n t a l f o r a n y p o s i t i v e n

Commented by malwaan last updated on 03/Jul/16

e^(π(√(163))) had been conjectured around 1914 by the indian mathematician Srinivasa Ramanujan . In May 1974 John Brillo of the university of Arizona managed to prove that e^(π(√(163))) =262 537 412 640 768 744

e^{π \sqrt{163}} h a d b e e n c o n j e c t u r e d a r o u n d

1914 b y t h e i n d i a n m a t h e m a t i c i a n

S r i n i v a s a R a m a n u j a n .

I n M a y 1974 J o h n B r i l l o o f t h e

u n i v e r s i t y o f A r i z o n a m a n a g e d t o

p r o v e t h a t

e^{π \sqrt{163}} = 262 537 412 640 768 744

Commented by Rasheed Soomro last updated on 03/Jul/16

New knowledqe for me.

N e w k n o w l e d q e f o r m e .

Commented by prakash jain last updated on 03/Jul/16

W. alpha clearly states that it is a trancendental number. It is almost an integer.

W . alpha clearly states that it is a trancendental

number .

It is almost an integer .

Commented by malwaan last updated on 04/Jul/16

I think you are right It is almost integer BUT my answer is from Crux Canadian mathematical Society. EUREKA NO. 4 June 1975

I t h i n k y o u a r e r i g h t

I t i s a l m o s t i n t e g e r

B U T m y a n s w e r i s f r o m

C r u x

C a n a d i a n m a t h e m a t i c a l S o c i e t y .

E U R E K A N O . 4 J u n e 1975

Answered by Temp last updated on 01/Jul/16

I just realised for a+b=c, a,b∈T c∈Z let a=π, b=n−a, n∈Z ∴a+b=π+n−π a+b=n

I just realised for a + b = c,

a, b \in T

c \in Z

let a = π, b = n - a, n \in Z

∴ a + b = π + n - π

a + b = n

Commented by Rasheed Soomro last updated on 01/Jul/16

π and n−π are derivations of same transcedental (π). Is there an example of two transcedentals which are not derivations of same transcedental and give the sum or product as whole number?

π a n d n - π a r e d e r i v a t i o n s o f s a m e

t r a n s c e d e n t a l (π) .

I s t h e r e a n e x a m p l e

o f t w o t r a n s c e d e n t a l s w h i c h a r e n o t

d e r i v a t i o n s o f s a m e t r a n s c e d e n t a l

a n d g i v e t h e s u m o r p r o d u c t a s w h o l e

n u m b e r ?

Commented by Temp last updated on 01/Jul/16

That is an interesting point which I am very interested to find

That is an interesting point which

I am very interested to find

Commented by Rasheed Soomro last updated on 01/Jul/16

An other question is: Are there two transcedentals, (which are not derived from same transcedental/s) whose sum/product is an algebraic number?

A n o t h e r q u e s t i o n i s :

A r e t h e r e t w o t r a n s c e d e n t a l s, (w h i c h a r e

n o t d e r i v e d f r o m s a m e t r a n s c e d e n t a l / s) w h o s e

s u m / p r o d u c t i s a n a l g e b r a i c n u m b e r ?

Commented by prakash jain last updated on 01/Jul/16

If sum/product is algebraic the second transcendental number can be derived from the first algebraic number by adding or subtracting the first transcental number. The problem statement implies that that second transcental and first transcendental numbers can be derived using the sum. so the two conditions are contrdictory t_1 +t_2 =n for some n∈Z (1) also second condition is t_1 ≠n−t_1 ∀n∈Z (2) (2) contradicts with (1) same applies for multiplication.

If sum / product is algebraic the second

transcendental number can be derived from

the first algebraic number by adding or subtracting

the first transcental number .

The problem statement implies that that

second transcental and first transcendental

numbers can be derived using the sum .

so the two conditions are contrdictory

t_{1} + t_{2} = n f o r s o m e n \in Z (1)

also second condition is

t_{1} \neq n - t_{1} \forall n \in Z (2)

(2) contradicts with (1)

same applies for multiplication .

Commented by Rasheed Soomro last updated on 03/Jul/16

Your comment is decisive!

Y o u r c o m m e n t i s d e c i s i v e!

Y o u r c o m m e n t i s d e c i s i v e!

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