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Prove ${e}^{x}$, $xinmathbb{Z}$ is irrational [duplicate]

Proving the irrationality of $e^n$(How to/Can I) show irrational numbers?Proof of $pi$+$e$ irrationalRational + irrational = always irrational?If $(n_k)$ is strictly increasing and $lim_{n to infty} n_k^{1/2^k} = infty$ show that $sum_{k=1}^{infty} 1/n_k$ is irrational$sqrt[3]{5}$ is irrationalIf the radioactive isotope strontium $240$ has a half life of $120$ years, how long until it decays to only $60%$ of its original radioactivity?How to prove $sqrt{18}$ is irrational without using proof by contradiction?Prove that $log_2 3$ is irrational!Prove $Big(frac{x+1}{x}Big)^{2/n}$ is irrational for $x,n in mathbb{N}, n>2$How do I prove if a sum of two specific irrational numbers is irrational?

0

1

$begingroup$

This question already has an answer here:

• 
  Proving the irrationality of $e^n$
  
  4 answers
  
  

I was thinking of random problems in my head when this came to mind. But I have absolutely no idea how to solve it, so I can't show my attempts. Please help me with this.

Prove $boldsymbol{e^x}$, $boldsymbol{xinmathbb{Z}}$ is irrational

exponential-function irrational-numbers

share|cite|improve this question

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

$endgroup$

marked as duplicate by Hans Engler, Eevee Trainer, Lord Shark the Unknown, Lee David Chung Lin, Cesareo Mar 14 at 9:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this math.stackexchange.com/questions/476939/… may interest you probaly
  $endgroup$
  – cand
  Mar 14 at 0:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not true for $x=0$
  $endgroup$
  – J. W. Tanner
  Mar 14 at 0:33
  

  
  
  
  

add a comment | 

0

1

$begingroup$

This question already has an answer here:

• 
  Proving the irrationality of $e^n$
  
  4 answers
  
  

I was thinking of random problems in my head when this came to mind. But I have absolutely no idea how to solve it, so I can't show my attempts. Please help me with this.

Prove $boldsymbol{e^x}$, $boldsymbol{xinmathbb{Z}}$ is irrational

exponential-function irrational-numbers

share|cite|improve this question

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

$endgroup$

marked as duplicate by Hans Engler, Eevee Trainer, Lord Shark the Unknown, Lee David Chung Lin, Cesareo Mar 14 at 9:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  this math.stackexchange.com/questions/476939/… may interest you probaly
  $endgroup$
  – cand
  Mar 14 at 0:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not true for $x=0$
  $endgroup$
  – J. W. Tanner
  Mar 14 at 0:33
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Proving the irrationality of $e^n$
  
  4 answers
  
  

I was thinking of random problems in my head when this came to mind. But I have absolutely no idea how to solve it, so I can't show my attempts. Please help me with this.

Prove $boldsymbol{e^x}$, $boldsymbol{xinmathbb{Z}}$ is irrational

exponential-function irrational-numbers

share|cite|improve this question

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

$endgroup$

share|cite|improve this question

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

share|cite|improve this question

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

edited Mar 14 at 1:42

Rócherz

2,9863821

edited Mar 14 at 1:42

Rócherz

2,9863821

asked Mar 14 at 0:14

Math BobMath Bob

11411

asked Mar 14 at 0:14

Math BobMath Bob

11411

1 Answer
1

active

oldest

votes

0

$begingroup$

It's known $e=e^1$ is transcendental. So for positive $n,$ if $e^n$ were rational, say $a/b,$ then it would be a root of $bx=a.$ Then $b cdot e^n=a,$ not possible for positive $n.$ Probably negative $n$ cn also be ruled out this way, but I'll leave that to you or other responders.

share|cite|improve this answer

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

$endgroup$

add a comment | 

0

$begingroup$

It's known $e=e^1$ is transcendental. So for positive $n,$ if $e^n$ were rational, say $a/b,$ then it would be a root of $bx=a.$ Then $b cdot e^n=a,$ not possible for positive $n.$ Probably negative $n$ cn also be ruled out this way, but I'll leave that to you or other responders.

share|cite|improve this answer

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

$endgroup$

add a comment | 

0

$begingroup$

It's known $e=e^1$ is transcendental. So for positive $n,$ if $e^n$ were rational, say $a/b,$ then it would be a root of $bx=a.$ Then $b cdot e^n=a,$ not possible for positive $n.$ Probably negative $n$ cn also be ruled out this way, but I'll leave that to you or other responders.

share|cite|improve this answer

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

$endgroup$

add a comment | 

$begingroup$

It's known $e=e^1$ is transcendental. So for positive $n,$ if $e^n$ were rational, say $a/b,$ then it would be a root of $bx=a.$ Then $b cdot e^n=a,$ not possible for positive $n.$ Probably negative $n$ cn also be ruled out this way, but I'll leave that to you or other responders.

share|cite|improve this answer

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

$endgroup$

share|cite|improve this answer

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415

answered Mar 14 at 0:23

coffeemathcoffeemath

2,9071415