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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?
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{such}\:\mathrm{that}\:\mathrm{two}\:\mathrm{trancendental} \\ $$$$\mathrm{numbers}\:\mathrm{add}\:\mathrm{or}\:\mathrm{multiply}\:\mathrm{to}\:\mathrm{give}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}? \\ $$
Commented by prakash jain last updated on 29/Jun/16
π is transcendental.  (1/π) is also transcendental.  π×(1/π)=1
$$\pi\:\mathrm{is}\:\mathrm{transcendental}. \\ $$$$\frac{\mathrm{1}}{\pi}\:\mathrm{is}\:\mathrm{also}\:\mathrm{transcendental}. \\ $$$$\pi×\frac{\mathrm{1}}{\pi}=\mathrm{1} \\ $$
Commented by Temp last updated on 29/Jun/16
what if in form ab, b≠a^(−1) ?
$${what}\:{if}\:{in}\:{form}\:{ab},\:{b}\neq{a}^{−\mathrm{1}} ? \\ $$
Commented by Rasheed Soomro last updated on 29/Jun/16
π×(2/π)=2, where (2/π)≠π^(-1)
$$\pi×\frac{\mathrm{2}}{\pi}=\mathrm{2},\:{where}\:\frac{\mathrm{2}}{\pi}\neq\pi^{-\mathrm{1}} \\ $$
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
$$\mathrm{The}\:\mathrm{way}\:\mathrm{I}\:\mathrm{see}\:\mathrm{it},\:\mathrm{the}\:\mathrm{multiplicitive} \\ $$$$\mathrm{form}\:{ab}={c},\:{a},{b}=\mathrm{transendental},\:{c}={integer} \\ $$$$\mathrm{it}\:\mathrm{seems}\:\mathrm{that}\:{ab}={c}\:\mathrm{only}\:\mathrm{holds}\:\mathrm{true}\:\mathrm{for}\:\mathrm{if} \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{share}\:\mathrm{a}\:\mathrm{relationship}\:\mathrm{of}: \\ $$$${a}=\frac{\mathrm{1}}{{b}} \\ $$$$\therefore{ab}=\frac{{b}}{{b}}={c} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{unsure}\:\mathrm{if}\:\mathrm{for}\:\mathrm{some}\:\mathrm{two}\:\mathrm{different} \\ $$$$\mathrm{transendental}\:\mathrm{numbers}\:{other}\:{than}\:\mathrm{when}\:{aa}^{−\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{exist}. \\ $$$$\mathrm{e}.\mathrm{g}.\:{e}\pi={z},\:{z}\notin\mathbb{Z} \\ $$
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)
$${n}\pi={k},\:{k}\in\mathbb{Z},\:\therefore{n}\neq\mathrm{1},\:{n}\:\mathrm{can}\:\mathrm{be}\:\mathrm{trancendental} \\ $$$$\therefore\pi=\frac{{k}}{{n}}\:=\:\boldsymbol{{false}} \\ $$$$\therefore\mathrm{Impossible}\:\mathrm{for}\:\mathrm{different}\:\mathrm{trancendentals} \\ $$$$\mathrm{to}\:\mathrm{equal}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{excluding}\:\mathrm{cases}\:\mathrm{like} \\ $$$$\pi×\frac{\mathrm{1}}{\pi}.\:\:\:\left({i}.{e}.\:{k}>\mathrm{1}\right) \\ $$
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?
$${Say}\:\pi,{e}\:{and}\:{other}\:{transidental}\:{numbers}, \\ $$$${which}\:{can}\:{not}\:{be}\:{derived}\:{from}\:{one}\:{another} \\ $$$${by}\:+,−,×,\boldsymbol{\div},{power},{root}\:{are}\:{said}\:{to}\:{be}\:\boldsymbol{{root}}-\boldsymbol{{transedentals}}. \\ $$$$\:\pi,−\pi,\pi^{−\mathrm{1}} ,{n}\pi,\frac{{m}\pi}{{n}}\left({m},{n}\in\mathbb{Z}\right)\:{are} \\ $$$${of}\:{same}\:{roots}.\boldsymbol{{e}}\:{and}\:\pi\:{are}\:\:{root}-{transcidentals}. \\ $$$${Now}\:{what}\:{you}\:{want}\:{to}\:{say},\:{can}\:{be}\:{said}\: \\ $$$$“{Sum}\:{or}\:{product}\:{of}\:{two}\:{transidentals} \\ $$$${of}\:{different}\:{roots}\:{can}'{t}\:{be}\:{whole}\:{numbers}'' \\ $$$${Am}\:{I}\:{right}? \\ $$
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.
$${ab}={n}\Rightarrow{b}=\frac{{n}}{{a}} \\ $$$$\mathrm{So}\:\mathrm{if}\:\mathrm{product}\:\mathrm{of}\:\mathrm{any}\:\mathrm{2}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{a}\:\mathrm{whole} \\ $$$$\mathrm{number}\:\mathrm{then}\:\mathrm{second}\:\mathrm{number}\:{b}={a}/{n}. \\ $$
Answered by malwaan last updated on 02/Jul/16
e^(π(√(163)))  is an integer  e^(π(√(163)))  = 262 537 412 640 768 744
$${e}^{\pi\sqrt{\mathrm{163}}} \:{is}\:{an}\:{integer} \\ $$$${e}^{\pi\sqrt{\mathrm{163}}} \:=\:\mathrm{262}\:\mathrm{537}\:\mathrm{412}\:\mathrm{640}\:\mathrm{768}\:\mathrm{744} \\ $$
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
$${According}\:{to}\:{WolframAlpha} \\ $$$${e}^{\pi\sqrt{\mathrm{163}}} \:=\mathrm{262537412640768743}.\mathrm{99999999999925007}…. \\ $$$${what}\:{do}\:{you}\:{think}? \\ $$$${very}\:{small}\:{error}\:{because}\:{of}\:{calculation}\:{rounding} \\ $$
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)
$$\mathrm{According}\:\mathrm{to}\:\mathrm{Wolfram}\:\mathrm{alpha}\:{e}^{\pi\sqrt{\mathrm{163}}} \:\mathrm{is}\: \\ $$$$\mathrm{a}\:\mathrm{transcendental}\:\mathrm{number}. \\ $$$$\mathrm{2}.\mathrm{6253}..×\mathrm{10}^{\mathrm{17}} \\ $$
Commented by Rasheed Soomro last updated on 03/Jul/16
According to wikiepedia  e^(π(√n))   is transcedental for any positive n
$${According}\:{to}\:{wikiepedia} \\ $$$${e}^{\pi\sqrt{{n}}} \:\:{is}\:{transcedental}\:{for}\:{any}\:{positive}\:{n} \\ $$
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
$${e}^{\pi\sqrt{\mathrm{163}}} \:{had}\:{been}\:{conjectured}\:{around} \\ $$$$\mathrm{1914}\:{by}\:{the}\:{indian}\:{mathematician} \\ $$$${Srinivasa}\:{Ramanujan}\:. \\ $$$${In}\:{May}\:\mathrm{1974}\:{John}\:{Brillo}\:{of}\:{the} \\ $$$${university}\:{of}\:{Arizona}\:{managed}\:{to} \\ $$$${prove}\:{that} \\ $$$${e}^{\pi\sqrt{\mathrm{163}}} \:=\mathrm{262}\:\mathrm{537}\:\mathrm{412}\:\mathrm{640}\:\mathrm{768}\:\mathrm{744} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 03/Jul/16
New knowledqe for me.
$${New}\:{knowledqe}\:{for}\:{me}. \\ $$
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.
$$\mathrm{W}.\:\mathrm{alpha}\:\mathrm{clearly}\:\mathrm{states}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{trancendental} \\ $$$$\mathrm{number}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{almost}\:\mathrm{an}\:\mathrm{integer}. \\ $$
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
$${I}\:{think}\:{you}\:{are}\:{right} \\ $$$${It}\:{is}\:\boldsymbol{{almost}}\:{integer} \\ $$$$\boldsymbol{{BUT}}\:\:{my}\:{answer}\:{is}\:{from} \\ $$$$\boldsymbol{{Crux}} \\ $$$$\boldsymbol{{Canadian}}\:\boldsymbol{{mathematical}}\:\boldsymbol{{Society}}. \\ $$$$\boldsymbol{{EUREKA}}\:\boldsymbol{{NO}}.\:\mathrm{4}\:\boldsymbol{{June}}\:\mathrm{1975} \\ $$
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
$$\mathrm{I}\:\mathrm{just}\:\mathrm{realised}\:\mathrm{for}\:{a}+{b}={c}, \\ $$$${a},{b}\in\mathbb{T} \\ $$$${c}\in\mathbb{Z} \\ $$$$\mathrm{let}\:{a}=\pi,\:{b}={n}−{a},\:{n}\in\mathbb{Z} \\ $$$$\therefore{a}+{b}=\pi+{n}−\pi \\ $$$${a}+{b}={n} \\ $$
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?
$$\pi\:{and}\:\:{n}−\pi\:{are}\:{derivations}\:{of}\:{same} \\ $$$${transcedental}\:\left(\pi\right). \\ $$$${Is}\:{there}\:{an}\:{example} \\ $$$${of}\:{two}\:{transcedentals}\:{which}\:{are}\:{not} \\ $$$${derivations}\:{of}\:{same}\:{transcedental} \\ $$$${and}\:{give}\:{the}\:{sum}\:{or}\:{product}\:{as}\:{whole} \\ $$$${number}? \\ $$
Commented by Temp last updated on 01/Jul/16
That is an interesting point which  I am very interested to find
$$\mathrm{That}\:\mathrm{is}\:\mathrm{an}\:\mathrm{interesting}\:\mathrm{point}\:\mathrm{which} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{very}\:\mathrm{interested}\:\mathrm{to}\:\mathrm{find} \\ $$
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?
$${An}\:{other}\:{question}\:{is}: \\ $$$${Are}\:\:{there}\:{two}\:{transcedentals},\:\left({which}\:{are}\right. \\ $$$$\left.{not}\:{derived}\:{from}\:{same}\:{transcedental}/{s}\right)\:{whose} \\ $$$${sum}/{product}\:{is}\:{an}\:\boldsymbol{{algebraic}}\:\boldsymbol{{number}}? \\ $$
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.
$$\mathrm{If}\:\mathrm{sum}/\mathrm{product}\:\mathrm{is}\:\mathrm{algebraic}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{transcendental}\:\mathrm{number}\:\mathrm{can}\:\mathrm{be}\:\mathrm{derived}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{algebraic}\:\mathrm{number}\:\mathrm{by}\:\mathrm{adding}\:\mathrm{or}\:\mathrm{subtracting} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{transcental}\:\mathrm{number}. \\ $$$$\mathrm{The}\:\mathrm{problem}\:\mathrm{statement}\:\mathrm{implies}\:\mathrm{that}\:\mathrm{that} \\ $$$$\mathrm{second}\:\mathrm{transcental}\:\mathrm{and}\:\mathrm{first}\:\mathrm{transcendental} \\ $$$$\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{derived}\:\mathrm{using}\:\mathrm{the}\:\mathrm{sum}. \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{two}\:\mathrm{conditions}\:\mathrm{are}\:\mathrm{contrdictory} \\ $$$${t}_{\mathrm{1}} +{t}_{\mathrm{2}} ={n}\:{for}\:{some}\:{n}\in\mathbb{Z}\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{also}\:\mathrm{second}\:\mathrm{condition}\:\mathrm{is} \\ $$$${t}_{\mathrm{1}} \neq{n}−{t}_{\mathrm{1}} \:\forall{n}\in\mathbb{Z}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{contradicts}\:\mathrm{with}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{same}\:\mathrm{applies}\:\mathrm{for}\:\mathrm{multiplication}. \\ $$
Commented by Rasheed Soomro last updated on 03/Jul/16
Your comment is decisive!
$${Your}\:{comment}\:{is}\:{decisive}! \\ $$

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