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Review your own paper in Mathematics

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Prove that $neg exists x$ such that $ lvert x-1rvert+lvert x+1 rvert

$|x-1| +|x+1| <1$Spivak 's Calculus (Chapter I, Problem 16c)Is $lvert a-brvertlelvert arvert+lvert brvert$ always true?Question about proving basic results of numbersIs this a valid proof of the squeeze theorem?How to solve $lvert{x}rvert - lvert{2+x}rvert = x$?Limits of transcendental functions by direct substitution (rigorous proofs)Why does $lvert x^2 rvert < 16$ imply $lvert x rvert < 4$?Is $lvertsqrt{x}rvert^2$ equal $lvert xrvert$?How to do $lvert x - 1 rvert < lvert x-3 rvert$Prove that for all $xinmathbb{R}enspace lvert xrvert + lvert x-6 rvertgeq 6$

5

0

$begingroup$

This question already has an answer here:

• 
  $|x-1| +|x+1| <1$
  
  4 answers
  
  

In Spivak Calculus, chapter 1 question 11 vi. asks the reader to find all numbers $x$ for which $lvert x-1rvert+lvert x+1 rvert <1$. Intuitively speaking, it is quite obvious that there is no number $x$ that would make this inequality true, and in checking a graph of the equation I am now certain of this. Despite the textbook not asking for a proof, I am curious as to how one would go about proving this rigorously? Would this require using multiple cases or is it possible without?

Thanks for any insight.

Note: Ideally any proof presented would not require calculus, as that is beyond the scope of the first chapter of the book, but I'd still be interested in any proof involving calculus as this is purely out of curiosity.

calculus algebra-precalculus inequality absolute-value

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edited Mar 13 at 6:58

Dastur

asked Mar 13 at 6:52

DasturDastur

1697

$endgroup$

marked as duplicate by Martin R, Gibbs, Vinyl_cape_jawa, Song, N. F. Taussig Mar 13 at 13:27

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.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of $|x-1| +|x+1| < 1$
  $endgroup$
  – Martin R
  Mar 13 at 8:31
  
  
  

  
  
  
  

add a comment | 

5

0

$begingroup$

This question already has an answer here:

• 
  $|x-1| +|x+1| <1$
  
  4 answers
  
  

In Spivak Calculus, chapter 1 question 11 vi. asks the reader to find all numbers $x$ for which $lvert x-1rvert+lvert x+1 rvert <1$. Intuitively speaking, it is quite obvious that there is no number $x$ that would make this inequality true, and in checking a graph of the equation I am now certain of this. Despite the textbook not asking for a proof, I am curious as to how one would go about proving this rigorously? Would this require using multiple cases or is it possible without?

Thanks for any insight.

Note: Ideally any proof presented would not require calculus, as that is beyond the scope of the first chapter of the book, but I'd still be interested in any proof involving calculus as this is purely out of curiosity.

calculus algebra-precalculus inequality absolute-value

share|cite|improve this question

edited Mar 13 at 6:58

Dastur

asked Mar 13 at 6:52

DasturDastur

1697

$endgroup$

marked as duplicate by Martin R, Gibbs, Vinyl_cape_jawa, Song, N. F. Taussig Mar 13 at 13:27

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.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of $|x-1| +|x+1| < 1$
  $endgroup$
  – Martin R
  Mar 13 at 8:31
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  $|x-1| +|x+1| <1$
  
  4 answers
  
  

In Spivak Calculus, chapter 1 question 11 vi. asks the reader to find all numbers $x$ for which $lvert x-1rvert+lvert x+1 rvert <1$. Intuitively speaking, it is quite obvious that there is no number $x$ that would make this inequality true, and in checking a graph of the equation I am now certain of this. Despite the textbook not asking for a proof, I am curious as to how one would go about proving this rigorously? Would this require using multiple cases or is it possible without?

Thanks for any insight.

Note: Ideally any proof presented would not require calculus, as that is beyond the scope of the first chapter of the book, but I'd still be interested in any proof involving calculus as this is purely out of curiosity.

calculus algebra-precalculus inequality absolute-value

share|cite|improve this question

edited Mar 13 at 6:58

Dastur

asked Mar 13 at 6:52

DasturDastur

1697

$endgroup$

share|cite|improve this question

edited Mar 13 at 6:58

Dastur

asked Mar 13 at 6:52

DasturDastur

1697

share|cite|improve this question

edited Mar 13 at 6:58

Dastur

asked Mar 13 at 6:52

DasturDastur

1697

asked Mar 13 at 6:52

DasturDastur

1697

asked Mar 13 at 6:52

DasturDastur

1697

3 Answers
3

active

oldest

votes

11

$begingroup$

Suppose there exists $x_0in Bbb R $ such that $lvert x_0-1rvert+lvert x_0+1 rvert <1$. Then
begin{align}|2|&=|x_0-1-(x_0+1)|\
&leq lvert x_0-1rvert+lvert -(x_0+1) rvert\&=lvert x_0-1rvert+lvert x_0+1 rvert\
& <1
end{align}
,a contradiction .

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answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

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add a comment | 

4

$begingroup$

Case 1: $x le -1$

$|x - 1| + |x + 1| = -(x - 1) + -(x + 1) = -2x ge 2$

Case 2 : $-1 le x le 1$

$|x - 1| + |x + 1| = -(x - 1) + (x + 1) = 2$

Case 3 : $1 le x$

$|x - 1| + |x + 1| = (x - 1) + (x + 1) = 2x ge 2$

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answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

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3

$begingroup$

Because by the triangle inequality $$|x+1|+|x-1|=|x+1|+|1-x|geq|x+1+1-x|=2geq1.$$

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answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

11

$begingroup$

Suppose there exists $x_0in Bbb R $ such that $lvert x_0-1rvert+lvert x_0+1 rvert <1$. Then
begin{align}|2|&=|x_0-1-(x_0+1)|\
&leq lvert x_0-1rvert+lvert -(x_0+1) rvert\&=lvert x_0-1rvert+lvert x_0+1 rvert\
& <1
end{align}
,a contradiction .

share|cite|improve this answer

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

$endgroup$

add a comment | 

11

$begingroup$

Suppose there exists $x_0in Bbb R $ such that $lvert x_0-1rvert+lvert x_0+1 rvert <1$. Then
begin{align}|2|&=|x_0-1-(x_0+1)|\
&leq lvert x_0-1rvert+lvert -(x_0+1) rvert\&=lvert x_0-1rvert+lvert x_0+1 rvert\
& <1
end{align}
,a contradiction .

share|cite|improve this answer

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

$endgroup$

add a comment | 

$begingroup$

Suppose there exists $x_0in Bbb R $ such that $lvert x_0-1rvert+lvert x_0+1 rvert <1$. Then
begin{align}|2|&=|x_0-1-(x_0+1)|\
&leq lvert x_0-1rvert+lvert -(x_0+1) rvert\&=lvert x_0-1rvert+lvert x_0+1 rvert\
& <1
end{align}
,a contradiction .

share|cite|improve this answer

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

answered Mar 13 at 6:59

Thomas ShelbyThomas Shelby

4,2692726

4

$begingroup$

Case 1: $x le -1$

$|x - 1| + |x + 1| = -(x - 1) + -(x + 1) = -2x ge 2$

Case 2 : $-1 le x le 1$

$|x - 1| + |x + 1| = -(x - 1) + (x + 1) = 2$

Case 3 : $1 le x$

$|x - 1| + |x + 1| = (x - 1) + (x + 1) = 2x ge 2$

share|cite|improve this answer

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

$endgroup$

add a comment | 

4

$begingroup$

Case 1: $x le -1$

$|x - 1| + |x + 1| = -(x - 1) + -(x + 1) = -2x ge 2$

Case 2 : $-1 le x le 1$

$|x - 1| + |x + 1| = -(x - 1) + (x + 1) = 2$

Case 3 : $1 le x$

$|x - 1| + |x + 1| = (x - 1) + (x + 1) = 2x ge 2$

share|cite|improve this answer

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

$endgroup$

add a comment | 

$begingroup$

Case 1: $x le -1$

$|x - 1| + |x + 1| = -(x - 1) + -(x + 1) = -2x ge 2$

Case 2 : $-1 le x le 1$

$|x - 1| + |x + 1| = -(x - 1) + (x + 1) = 2$

Case 3 : $1 le x$

$|x - 1| + |x + 1| = (x - 1) + (x + 1) = 2x ge 2$

share|cite|improve this answer

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

$endgroup$

share|cite|improve this answer

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

answered Mar 13 at 7:24

DanielVDanielV

18.1k42755

3

$begingroup$

Because by the triangle inequality $$|x+1|+|x-1|=|x+1|+|1-x|geq|x+1+1-x|=2geq1.$$

share|cite|improve this answer

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

3

$begingroup$

Because by the triangle inequality $$|x+1|+|x-1|=|x+1|+|1-x|geq|x+1+1-x|=2geq1.$$

share|cite|improve this answer

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

$begingroup$

Because by the triangle inequality $$|x+1|+|x-1|=|x+1|+|1-x|geq|x+1+1-x|=2geq1.$$

share|cite|improve this answer

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

share|cite|improve this answer

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200

answered Mar 13 at 8:51

Michael RozenbergMichael Rozenberg

108k1895200