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

Question

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 $27$.

@lulu as in another question. But here $1<\sqrt{2}<e<9$ so it is not so easy — Henry, Jan 04 '23 at 12:38
@lulu true but wont that work only for both quantities grater than $e$ or both less than $e$. Here $9>e, \sqrt{2}<e$. — Ekaveera Gouribhatla, Jan 04 '23 at 12:38
Using a calculator to cheat a little, we see that $9^{\sqrt2} \approx 22.36$ while $\sqrt{2}^9\approx 22.63$. So it is, for instance, not possible to solve this by wedging an integer between them. If you square them you get $500.04$ versus $512$. Maybe those are easier to compare? — Arthur, Jan 04 '23 at 12:41
If we use $\sqrt2<\frac{17}{12}$, then it's enough to show $3^{17}<2^{27}$. — Milten, Jan 04 '23 at 12:56
@lulu What you said is correct, with simple manipulation we can use calculus as shown by two authors below :) — Ekaveera Gouribhatla, Jan 04 '23 at 14:39

score 18 · Answer 1 · answered Jan 04 '23 at 14:26

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.

Henry · Accepted Answer · 2023-01-04T13:31:45.640

12

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

edited Jan 04 '23 at 13:31

answered Jan 04 '23 at 12:53

Henry

157,058

2

My solution as well. Don't know why the downvote – Milten Jan 04 '23 at 12:58
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@Milten, Maybe the downvoter didn't like the extraction of a cube root of a nine digit number sans calculator. – Peter Phipps Jan 04 '23 at 13:04
Boohhhh...many times I don't understand!+1 – Sebastiano Jan 04 '23 at 13:07
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@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). – Milten Jan 04 '23 at 13:11
Mine as well, +1. I just came up to prove $9^{17}<2^{54}$ ... – Michael Hoppe Jan 04 '23 at 13:18
3

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 – Henry Jan 04 '23 at 13:30
Thank u very much Henry :) – Ekaveera Gouribhatla Jan 04 '23 at 13:39

achille hui · Answer 3 · 2023-01-04T14:22:41.107

12

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.

edited Jan 04 '23 at 14:22

answered Jan 04 '23 at 14:15

achille hui

122,701

(+1). Actually our proofs are similar. – River Li Jan 04 '23 at 14:32
Nice missed out this simple trick..Tq – Ekaveera Gouribhatla Jan 04 '23 at 14:37

score 4 · Answer 4 · answered Jan 04 '23 at 14:56

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

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

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

Bob Dobbs · Answer 5 · 2023-01-04T18:38:32.557

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:

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

score 2 · Answer 6 · answered Jan 10 '23 at 21:24

2

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

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

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

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

answered Jan 10 '23 at 21:24

albert chan

2,114

It is nice. (+1) – River Li Jan 12 '23 at 00:55

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

6 Answers6