Tag for problems about comparing explicitly given numbers, often by hand calculation only.
Questions tagged [number-comparison]
214 questions
63
votes
18 answers
Which is larger? $20!$ or $2^{40}$?
Someone asked me this question, and it bothers the hell out of me that I can't prove either way. I've sort of come to the conclusion that 20! must be larger, because it has 36 prime factors, some of which are significantly larger than 2, whereas…
Alec
- 4,184
43
votes
4 answers
Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator
Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator. I did in the following way. Are there other ways? Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then, $$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$ Now, $$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad…
mathlove
- 155,494
36
votes
7 answers
Prove $7^{71}>75^{32}$
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$ I did the first question:…
lsr314
- 16,098
35
votes
10 answers
Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater?
Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$? I want to know if my proof is correct... \begin{align} \sqrt[8]{8!} &< \sqrt[9]{9!} \\ (8!)^{(1/8)} &< (9!)^{(1/9)} \\ (8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\ (8!)^{(9/72)} - (9!)^{8/72} &<…
Sonia f
- 369
30
votes
4 answers
Prove that $\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$
show that $$\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$$ and I found $$LHs-RHS=0.017\cdots$$ I have post this interesting problem Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ can someone suggest any other nice method? Thank you everyone.
math110
- 95,328
- 17
- 156
- 531
27
votes
13 answers
Which is greater, $98^{99} $ or $ 99^{98}$?
Which is greater, $98^{99} $ or $ 99^{98}$? What is the easiest method to do this which can be explained to someone in junior school i.e. without using log tables. I don't think there is an elementary way to do this. The best I could find was on…
Bhaskar Vashishth
- 11,756
26
votes
8 answers
Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$
How to prove the following trignometric identity? $$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$ Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 +…
Parth Thakkar
- 4,618
25
votes
7 answers
Show by hand : $e^{e^2}>1000\phi$
Problem: Show by hand without any computer assistance: $$e^{e^2}>1000\phi,$$ where $\phi$ denotes the golden ratio $\frac{1+\sqrt{5}}{2} \approx 1.618034$ . I come across this limit showing: $$\lim_{x\to…
Barackouda
- 3,915
- 3
- 13
- 43
24
votes
7 answers
Simple proof that $8\left(\frac{9}{10}\right)^8 > 1$
This question is motivated by a step in the proof given here . $\begin{align*} 8^{n+1}-1&\gt 8(8^n-1)\gt 8n^8\\ &=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\ &\geq (n+1)^8\left(8\left(\frac{9}{10}\right)^8\right)\\ &\gt (n+1)^8 .…
JasonMond
- 4,114
21
votes
7 answers
Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator
Is there a slick elementary way of proving that $3^{(3^4)}>4^{(4^3)}$ without using a calculator? Here is what I was thinking: $$4^4=256>243=3^5,$$ hence…
A. Goodier
- 11,362
- 7
- 35
- 54
20
votes
2 answers
Elegantly Proving that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\frac12$
$\qquad$ How could we prove, without the aid of a calculator, that $~\sqrt[5]{12}~-~\sqrt[12]5~>~\dfrac12$ ? I have stumbled upon this mildly interesting numerical coincidence by accident, while pondering on another curios approximation, related to…
Lucian
- 49,592
19
votes
9 answers
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
How can you determine which one of these numbers is bigger (without calculating): $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
user2637293
- 1,796
18
votes
5 answers
Without using a calculator and logarithm, which of $100^{101} , 101^{100}$ is greater?
Which of the following numbers is greater? Without using a calculator and logarithm. $$100^{101} , 101^{100}$$ My try : $$100=10^2\\101=(100+1)=(10^2+1)$$ So : $$100^{101}=10^{2(101)}\\101^{100}=(10^2+1)^{100}=10^{2(100)}+N$$ Now what ?
Almot1960
- 5,162
16
votes
3 answers
Compare $\log_5(\log_4 3)$ and $\log_6(\log_6 3)$
Let $a=\log_5(\log_4 3)$ and $b=\log_6(\log_6 3)$ . How to compare the two numbers if numerical solution is not used? The following is my thoughts, but it did not work. From the question, I get $5^a=\log_4 3$ and $6^b=\log_6 3$ . As $1 >\log_4 3 >…
Elysia
- 171
16
votes
7 answers
Show that $\sqrt{10}+\sqrt{26}+\sqrt{17}+\sqrt{37} \gt \sqrt{341}$
Show that $\sqrt{10}+\sqrt{26}+\sqrt{17}+\sqrt{37} \gt \sqrt{341}$ . This is inspired by Showing $x+y>z$, where $x=\sqrt{10}+\sqrt{26}$, $y=\sqrt{17}+\sqrt{37}$, and $z=\sqrt{323}$. Is my idea corect? , where the 341 is replaced by 323. In that…
marty cohen
- 110,916