Ygtjki

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Quick but not simple question. $2^sqrt2$ or e, which is greater?

Which of these numbers is greater: $sqrt[5]5$ or $sqrt[4]4$?Inequality through calculusWhich is greater, $300 !$ or $(300^300)^frac 12$?Hausdorff metric and Vietoris topologyClosed and Bounded but not compactWhich coefficient is greater?Calculating square roots using the recurrence $x_n+1 = frac12 left(x_n + frac2x_nright)$Trigonometric inequality - x domain$e^left(pi^(e^pi)right);$ or $;pi^left(e^(pi^e)right)$. Which one is greater than the other?Prove by contradiction (not using a calculator) that $sqrt6 + sqrt2 < sqrt15$?

5

2

$begingroup$

$2^sqrt2$ vs $e$, which is greater?

$(2^sqrt2)^sqrt2 = 4quad $ & $quad e^sqrt2$ = ?

$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$

I tried but can't induce comparable form.

Is anybody know how to prove it?

real-analysis analysis inequality

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question?
  $endgroup$
  – Michael Rozenberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2^√2 > e or 2^√2 < e ? is my question
  $endgroup$
  – J.Bo
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the solution to $2^x=e$ is $x=frac 1ln 2approx 1.4427>sqrt 2$. Of course, numerical computation is involved in that.
  $endgroup$
  – lulu
  yesterday
  

  
  
  
  


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  3
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with using a calculator?
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's simple but good idea. THX
  $endgroup$
  – J.Bo
  yesterday
  

  
  
  
  

 | 
show 3 more comments

5

2

$begingroup$

$2^sqrt2$ vs $e$, which is greater?

$(2^sqrt2)^sqrt2 = 4quad $ & $quad e^sqrt2$ = ?

$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$

I tried but can't induce comparable form.

Is anybody know how to prove it?

real-analysis analysis inequality

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question?
  $endgroup$
  – Michael Rozenberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2^√2 > e or 2^√2 < e ? is my question
  $endgroup$
  – J.Bo
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the solution to $2^x=e$ is $x=frac 1ln 2approx 1.4427>sqrt 2$. Of course, numerical computation is involved in that.
  $endgroup$
  – lulu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with using a calculator?
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's simple but good idea. THX
  $endgroup$
  – J.Bo
  yesterday
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

$2^sqrt2$ vs $e$, which is greater?

$(2^sqrt2)^sqrt2 = 4quad $ & $quad e^sqrt2$ = ?

$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$

I tried but can't induce comparable form.

Is anybody know how to prove it?

real-analysis analysis inequality

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

edited yesterday

YuiTo Cheng

2,1362837

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

asked yesterday

J.BoJ.Bo

456

4 Answers
4

active

oldest

votes

8

$begingroup$

Instead of comparing $2^sqrt 2$ and $e$, let's raise both to $sqrt 2$ and compare $2^2$ and $e^sqrt 2$:
$$
e^sqrt 2 > 2.7^1.4 approx 4.017068799 > 4 = 2^2
$$
Or use that
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
with $x=1.41$ and get
$$
e^sqrt 2 > e^1.41 > 4.03594 > 4
$$
In fact, $e^sqrt 2 approx 4.113250377 > 4$.

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

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• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand how you can estimate $2.7^1.4$, $e^1.41$, and $e^sqrt 2$ without a calculator. And if you have a calculator, why not find $2^sqrt 2$ from the beginning?
  $endgroup$
  – Teepeemm
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm, you're right. The best approach is the second one, with a polynomial. Unfortunately, it's of degree $4$ and you have to use two decimals in $x=1.41$.
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  

add a comment | 

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

5

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.

It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. It's a little trickier to show that $ln(2)lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice solution because this does not require a calculator.
  $endgroup$
  – quarague
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$...
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  


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  $begingroup$
  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


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  $begingroup$
  I see. It's not your fault. Corrected. It's a careless typo.
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


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  $begingroup$
  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


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  $begingroup$
  Yeah, it's exactly the summary of the first answer.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

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8

$begingroup$

Instead of comparing $2^sqrt 2$ and $e$, let's raise both to $sqrt 2$ and compare $2^2$ and $e^sqrt 2$:
$$
e^sqrt 2 > 2.7^1.4 approx 4.017068799 > 4 = 2^2
$$
Or use that
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
with $x=1.41$ and get
$$
e^sqrt 2 > e^1.41 > 4.03594 > 4
$$
In fact, $e^sqrt 2 approx 4.113250377 > 4$.

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand how you can estimate $2.7^1.4$, $e^1.41$, and $e^sqrt 2$ without a calculator. And if you have a calculator, why not find $2^sqrt 2$ from the beginning?
  $endgroup$
  – Teepeemm
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm, you're right. The best approach is the second one, with a polynomial. Unfortunately, it's of degree $4$ and you have to use two decimals in $x=1.41$.
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  

add a comment | 

8

$begingroup$

Instead of comparing $2^sqrt 2$ and $e$, let's raise both to $sqrt 2$ and compare $2^2$ and $e^sqrt 2$:
$$
e^sqrt 2 > 2.7^1.4 approx 4.017068799 > 4 = 2^2
$$
Or use that
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
with $x=1.41$ and get
$$
e^sqrt 2 > e^1.41 > 4.03594 > 4
$$
In fact, $e^sqrt 2 approx 4.113250377 > 4$.

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand how you can estimate $2.7^1.4$, $e^1.41$, and $e^sqrt 2$ without a calculator. And if you have a calculator, why not find $2^sqrt 2$ from the beginning?
  $endgroup$
  – Teepeemm
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Teepeemm, you're right. The best approach is the second one, with a polynomial. Unfortunately, it's of degree $4$ and you have to use two decimals in $x=1.41$.
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Instead of comparing $2^sqrt 2$ and $e$, let's raise both to $sqrt 2$ and compare $2^2$ and $e^sqrt 2$:
$$
e^sqrt 2 > 2.7^1.4 approx 4.017068799 > 4 = 2^2
$$
Or use that
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
with $x=1.41$ and get
$$
e^sqrt 2 > e^1.41 > 4.03594 > 4
$$
In fact, $e^sqrt 2 approx 4.113250377 > 4$.

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

$endgroup$

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

answered yesterday

lhflhf

167k11172403

answered yesterday

lhflhf

167k11172403

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

5

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.

It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. It's a little trickier to show that $ln(2)lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice solution because this does not require a calculator.
  $endgroup$
  – quarague
  yesterday
  

  
  
  
  

add a comment | 

5

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If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.

It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. It's a little trickier to show that $ln(2)lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

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answered yesterday

Barry CipraBarry Cipra

60.5k655128

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  Nice solution because this does not require a calculator.
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  – quarague
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.

It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. It's a little trickier to show that $ln(2)lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

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answered yesterday

Barry CipraBarry Cipra

60.5k655128

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share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

answered yesterday

Barry CipraBarry Cipra

60.5k655128

answered yesterday

Barry CipraBarry Cipra

60.5k655128

0

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$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

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answered yesterday

FelorFelor

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  It's $2^sqrt2$...
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  – YuiTo Cheng
  yesterday
  

  
  
  
  


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  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
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  – Felor
  yesterday
  
  

  
  
  
  


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  I see. It's not your fault. Corrected. It's a careless typo.
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  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


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  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
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  – Felor
  yesterday
  
  

  
  
  
  


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  Yeah, it's exactly the summary of the first answer.
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  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$...
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. It's not your fault. Corrected. It's a careless typo.
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's exactly the summary of the first answer.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this answer

answered yesterday

FelorFelor

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New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

FelorFelor

1