Raging Roots

Algebra Level 3

$\Large \sqrt[1]{1} \qquad \sqrt[2]{2} \qquad \sqrt[3]{3} \qquad \sqrt[4]{4}$

Without using a calculator, compare the numbers above. Which of the following options is true?

\sqrt[4]{4} = \sqrt[2]{2} > \sqrt[3]{3} > \sqrt[1]{1}

\sqrt[2]{2} = \sqrt[4]{4} > \sqrt[3]{3} > \sqrt[1]{1}

\sqrt[1]{1} > \sqrt[2]{2} = \sqrt[3]{3} = \sqrt[4]{4}

\sqrt[3]{3} > \sqrt[4]{4} > \sqrt[2]{2} > \sqrt[1]{1}

\sqrt[4]{4} > \sqrt[3]{3} > \sqrt[2]{2} > \sqrt[1]{1}

\sqrt[1]{1} = \sqrt[2]{2} = \sqrt[3]{3} = \sqrt[4]{4}

\sqrt[3]{3} > \sqrt[2]{2} = \sqrt[4]{4} > \sqrt[1]{1}

\sqrt[1]{1} > \sqrt[2]{2} > \sqrt[3]{3} > \sqrt[4]{4}

5 solutions

Abdur Rehman Zahid
May 20, 2016

Let $\square$ represent the inequality/equality signs.Then: $\sqrt[1]{1}\;\square \;\sqrt[2]{2}\;\square \;\sqrt[3]{3}\;\square\;\sqrt[4]{4}$ Since all the terms are positive,raising them all to the $12^{\text{th}}$ power preserves their ordering.Doing so yields: $(\sqrt[1]{1})^{12}\;\square (\sqrt[2]{2})^{12}\;\square \;(\sqrt[3]{3})^{12}\;\square\; (\sqrt[4]{4})^{12}\\ 1\;\square \;64\;\square\; 81\;\square \;64$ Since $64=64$ ,we get that $\sqrt[4]{4}=\sqrt[2]{2}$ .Also,since $81>64$ and $64>1$ we know that $\sqrt[3]{3}>\sqrt[2]{2}=\sqrt[4]{4}$ and that $\sqrt[2]{2}=\sqrt[4]{4}>1$ .Combing these inequalities, we get: $\boxed{\sqrt[3]{3}>\sqrt[4]{4}=\sqrt[2]{2}>\sqrt[1]{1}}$

This is the best of all :)(+1)

Rohit Udaiwal - 5 years ago

Same solution :) , +1

Novril Razenda - 4 years, 11 months ago

Akash Patalwanshi
May 19, 2016

Relevant wiki: Exponential Inequalities

$\large 1^1 = 1, \sqrt{2} = 2^{\frac{1}{2}} = 4^{\frac{1}{4}},$ $\large (2^{\frac{1}{2}})^2 = 2 < (3^{\frac{1}{3}})^2= 3^{\frac{2}{3}}$ $→2^{\frac{1}{2}} < 3^{\frac{1}{3}}$

that is, $3^{\frac{1}{3}} > 2^{\frac{1}{2}}$

Thus, $\large 3^{\frac{1}{3}} > 2^{\frac{1}{2}} = 4^{\frac{1}{4}} > 1^1$

We get an accurate answer if we raise to the power of 12.

Samara Simha Reddy - 5 years ago

Or power of 6

Hung Woei Neoh - 5 years ago

Yeah even power 6 will be great!

Samara Simha Reddy - 5 years ago

Wait, how do you even know that $2 < 3^{\frac{2}{3}}$ ???

Hung Woei Neoh - 5 years ago

Let's compare $2^{\frac{1}{2}}$ and $3^{\frac{1}{3}}$ .Making the denominators of the exponents equal,we get $2^{\frac{3}{6}}$ and $3^{\frac{2}{6}}$ .Clearly, $3^{\frac{2}{6}}>2^{\frac{3}{6}}$ .As the finishing step,raise both sides by $\dfrac{6}{3}$ to get $3^{\frac{2}{6}\cdot\frac{6}{3}}>2^{\frac{3}{6}\cdot\frac{6}{3}}$ . $\therefore 3^{\frac{2}{3}}>2$

Rohit Udaiwal - 5 years ago

Oh, thanks for explaning

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – :-) if we provide details to each steps solution would get long and doesn't look cool. The reason for $2 < 3^{\frac{2}{3}}$ is explained very clearly by Rohit.

akash patalwanshi - 5 years ago

@Akash Patalwanshi – You should probably explain every step clearly,though not in detail,but a slight sketch will work.A role of a solution is to explain the steps in short yet fluent.Nevertheless,good job!

Rohit Udaiwal - 5 years ago

@Rohit Udaiwal – Yes. I am beginner in Latex typing. For typing above solution, I had spend more than 15 minutes :-(. Thats the main reason otherwise I could provide the solution with full details.

akash patalwanshi - 5 years ago

@Akash Patalwanshi – Oh no problem,I remember taking 30 minutes to write a solution for this problem.Lol :D

Rohit Udaiwal - 5 years ago

@Rohit Udaiwal – I remember writing a solution which took me 45 minutes ti write.Deleted it though because it was brute force.

Abdur Rehman Zahid - 5 years ago

@Akash Patalwanshi – No worries. It takes time to learn and master $\LaTeX$ writing. I learn faster because I have taken programming courses. It's good that you try to learn and get used to it.

I have written extremely long solutions, like this , this , this and this . Actually, as a matter of fact, my solutions are usually very long. I prefer it that way though, it makes my solution clearer. These 4 are the extra long ones, which took me about an hour for each of them.

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – Thanks. I will try to become perfect in latex typing soon. :-)

akash patalwanshi - 5 years ago

Hung Woei Neoh
May 20, 2016

$\sqrt[1]{1} = 1\\ \sqrt[2]{2} = 2^{\frac{1}{2}}\\ \sqrt[3]{3} = 3^{\frac{1}{3}}\\ \sqrt[4]{4} = (2^2)^{\frac{1}{4}} = 2^{\frac{1}{2}}$

From here, we know that $\sqrt[2]{2}=\sqrt[4]{4}$

Now, we need to compare three numbers. We take the sixth power of these numbers:

$1^6 = 1\\ (2^{\frac{1}{2}})^6 = 2^3 = 8\\ (3^{\frac{1}{3}})^6 = 3^2 = 9$

We know that:

$9 > 8 > 1\\ 9^{\frac{1}{6}} > 8^{\frac{1}{6}} > 1^{\frac{1}{6}}\\ 3^{\frac{1}{3}} > 2^{\frac{1}{2}} > 1\\ \boxed{\sqrt[3]{3} > \sqrt[2]{2} = \sqrt[4]{4} > \sqrt[1]{1}}$

Akash Shukla
May 20, 2016

This problem is of type $y=x^\frac{1}{x}$ . , which has its maximum value ay x=e. $y'= \frac{x^\frac{1}{x}*(1-lnx)}{x^2}$ From $0<x<e$ , it is increasing. From $e<x<infinite$ it is decreasing. So $1^1 < 2^1/2$ . $3^1/3 > 4^1/4 =2^1/2 > 1^1$

Chintan Raiyani
Jun 3, 2016

$\sqrt[1]{1} = \sqrt[12]{1} \\ \sqrt[2]{2} = \sqrt[12]{64} \\ \sqrt[3]{3} = \sqrt[12]{81} \\ \sqrt[4]{4} = \sqrt[12]{64} \\ (\sqrt[12]{81} = \sqrt[3]{3}) > (\sqrt[12]{64} = \sqrt[2]{2}) = (\sqrt[12]{64} = \sqrt[4]{4}) > (\sqrt[12]{1} = \sqrt[1]{1})$

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