# Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$

Without calculator prove that $$9^{\sqrt{2}} < \sqrt{2}^9$$.

My effort: I tried using the fact $$9^{\sqrt{2}}<9^{1.5}=27.$$

Also We have $$512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$$. But both are below $$27$$.

• How do you know that $2^{4.5}\lt 27$? Commented Jan 4, 2023 at 12:30
• @JohnDouma its very simple $512<729 \Rightarrow 2^9<27^2$ Commented Jan 4, 2023 at 12:32
• Then you should add that to your proof. Commented Jan 4, 2023 at 12:33
• Absolutely right, I have deleted that comment.
– lulu
Commented Jan 4, 2023 at 12:38
• If we use $\sqrt2<\frac{17}{12}$, then it's enough to show $3^{17}<2^{27}$. Commented Jan 4, 2023 at 12:56

Remark: @achille hui posted a similar proof. But we got them independently.

The desired inequality is written as $$3^{2\sqrt 2} < (2\sqrt 2)^3$$ or $$2\sqrt 2\, \ln 3 < 3\ln (2\sqrt 2)$$ or $$\frac{\ln 3}{3} < \frac{\ln(2\sqrt 2)}{2\sqrt 2}. \tag{1}$$

Let $$f(x) := \frac{\ln x}{x}.$$ We have $$f'(x) = \frac{1 - \ln x}{x^2}.$$ Thus, $$f'(x) < 0$$ on $$(\mathrm{e}, \infty)$$.

Since $$\mathrm{e} < 2\sqrt 2 < 3$$, (1) is true.

We are done.

What you want is a suitable rational upper bound to $$\sqrt{2} \approx 1.4142$$ such as $$\frac{17}{12} \approx 1.4167$$ which you can find for example with continued fractions. As a check, $$\left(\frac{17}{12}\right)^2=\frac{289}{144}>2$$.

Then you can say $$9^{\sqrt{2}} \lt 9^{17/12}=3^{17/6} = 129140163^{1/6} < 134217728^{1/6} =2^{27/6} =2^{9/2} = \sqrt{2}^9$$

• My solution as well. Don't know why the downvote Commented Jan 4, 2023 at 12:58
• @Milten, Maybe the downvoter didn't like the extraction of a cube root of a nine digit number sans calculator. Commented Jan 4, 2023 at 13:04
• Boohhhh...many times I don't understand!+1 Commented Jan 4, 2023 at 13:07
• @PeterPhipps I can see that, though the cube root argument is artificial. It boils down to calculating $3^{17}$ and $2^{37}$, which is tedious for sure, but should be doable in a realistic amount of time (e.g. using exponentiation by squaring or similar). Commented Jan 4, 2023 at 13:11
• In reality the process was $2^{27}$ to get $134217728$ rather than the other way round, but then reversed to get a line of $<$s and $=$s Commented Jan 4, 2023 at 13:30

The given inequality can be transformed to various equivalent forms: \begin{align} 9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9 &\iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3} \iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3\\ &\iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}\end{align}

Notice $$3 > 2\sqrt{2} \sim 2.828 > e \sim 2.718$$ and the function $$x^{\frac1x}$$ is strictly decreasing for $$x > e$$ (standard calculus exercise). The rightmost "inequality" is true and hence the original "inequality" $$9^{\sqrt{2}} < \sqrt{2}^9$$ is true.

• (+1). Actually our proofs are similar. Commented Jan 4, 2023 at 14:32
• Nice missed out this simple trick..Tq Commented Jan 4, 2023 at 14:37

Using Henry's approach, but with different estimates. If you know that the first three digits of $$\ \sqrt{2}\$$ is $$\ 1.41,\$$ then you have: $$\ 1.40 < \sqrt{2} < 1.42857\ldots,\$$ i.e.,

$$\ \frac{7}{5} < \sqrt{2} < \frac{10}{7}.$$

So,

$$\left( 9^{\sqrt{2}} \right)^7 < \left( 9^{ \frac{10}{7}} \right)^7 = 9^{10} = {81}^5 = \left( 80+1 \right)^5$$

$$= {80}^5 + 5\times {80}^4 + 10\times {80}^3 + 10\times {80}^2 + 5\times 80 + 1$$

$$< {80}^5 + 6 \times {80}^4 = 86 \times {80}^4 < 100 \times{80}^4$$

$$=100\times 8^4 \times {10}^4 < 4,100,000,000 = 4.1 \times 10^{9},$$

whereas

$$\left(\left(\sqrt{2}\right)^9\right)^7 = \left(\sqrt{2}\right)^{63} = 2^{31}\times \sqrt{2} > 1.4 \times 2^{31} = \frac{7}{2} \times 2^{31}$$

$$= 7 \times 2^{30} = 7 \times \left( 2^{10} \right)^3 = 7 \times 1024^{3} > 7 \times 1000^3 = 7 \times 10^9.$$

It boils down to the comparison of some powers of $$2$$ and $$3$$, if I didn't do any mistake, in the following way: $$9^{\sqrt{2}}<\sqrt{2}^9$$ if $$9<2^{\frac{9}{2\sqrt{2}}}$$, by using the fact that $$\sqrt{2}<1.415$$, if $$9<2^{\frac{9}{2\times 1.415}}$$ if $$9<2^{3.18}$$ if $$3^{100}<2^{159}$$ which is true since:

1. $$2^{159}>\frac{1}{2}(1,02\times 10^{3})^{16}>\frac{1.32\times10^{48}}{2}=6,6\times10^{47}$$.
2. $$3^{100}<(6\times 10^{4})^{10}=3^{10}\times 2^{10}\times 10^{40}<6\times10^{4}\times 1025\times 10^{40}=6,15\times 10^{47}$$

$$(\sqrt{2})^9 \quad vs.\quad 9^\sqrt{2}\qquad$$// square both side

$$2^9=512 \quad vs.\quad 3^{4\sqrt{2}} \;≈\; 3^{5.657} \;<\; 3^5\,3^\frac{2}{3}$$

$$\displaystyle 3^5\,3^\frac{2}{3} = 243×2×\sqrt[3]{1+\frac{1}{8}} \;<\;486×\left(1+\frac{1}{3×8}\right) = 506.25$$

$$→ (\sqrt{2})^9 \;>\; 9^\sqrt{2}\qquad$$

• It is nice. (+1) Commented Jan 12, 2023 at 0:55