# How would you prove that $2^{50} < 3^{33}$ without directly calculating the values [closed]

Could you generalise the question and get something along the lines of $n^{50} < (n+1)^{33}$ ?

## closed as unclear what you're asking by Did, Davide Giraudo, Lord Shark the Unknown, Yujie Zha, user223391 May 14 '17 at 23:41

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• Is it so? 😅 Let me check through calculator first. – Fawad May 11 '17 at 11:44
• My friend Ruby confirms the statement. – mvw May 11 '17 at 11:45
• Your proposed generalization is false, because $\lim_{n\to\infty} \frac{n^{50}}{(n+1)^{33}} =\lim_{n\to\infty}\frac{n^{50}}{n^{33}}=\infty$. For large enough $n$ the lower-order terms in the denominator are insignificant. – MJD May 11 '17 at 11:53
• It seems several users are considering that $$2^{11}<3^7$$ is a basic fact that can be used as a building block in a proof that $$2^{50}<3^{33}$$ Why is that so? I have no problem with asking my students to be aware of the fact that $$2^{10}>10^3$$ because of the central role of $2^{10}=1024$ in computer science but frankly, are we supposed to know that $$3^7>2048\ ?$$ – Did May 11 '17 at 12:13
• In general you can use low precision real numbers, if that is allowed. You use a variation of the subtractive Euclidean algorithm. Here you use divison instead. So you start off with $2<3$ and divide by the smaller number to get $(2,1.5)$. Now the smaller number is $1.5$ so divide to get $(1.33,1.5)$ and so on. At each stage you keep track of the divisions getting bounds like $2^1<3^1$, $2^2>3^1$, $2^3<3^2$, and so on. The ratio of the exponents are convergents to continued fraction for $\log(3/2)$. You stop when you get bounds close enough to locate $50/33$ above or below the bounds. – Somos May 11 '17 at 21:32

$2^{11} = 2048 < 2187 = 3^7$ implies $2^{44} < 3^{28}$.

$2^{6} = 64 < 243 = 3^5$ implies $2^{50} < 3^{33}$.

The methods given in the other answers are all very complicated. Furthermore, as Did points out in a comment they all depend on facts which are not in principle any less complex than the statement that is to be proved. The following method is quite simple and satisfies the request with no advanced theory whatever and “without calculating the values” as required:

Take a heap of red beans of size $2^{50}$ and a heap of navy beans of size $3^{33}$. Repeatedly remove one bean from each pile until the red pile is exhausted. At that point some navy beans will remain and the claim is proved.

• The mean volume of a single bean is around $.2$ cm${}^3$ hence the volume of the smaller heap of beans should be around $.2$ km${}^3$. The larger heap is roughly $5$ times this, that is, a cube with side roughly equal to $1$ km. Good luck with the procedure... – Did May 14 '17 at 21:28

The easiest way is to derive the inequality from smaller ones. First you compute $$2^3 < 3^2 \tag 1$$ and $$2^{11} < 3^7 \tag 2$$ then you can combine these by multiplying inequalities (1) twice and (2) once: $$2^{17} < 3^{11} \tag 3$$ Cube this to get $$2^{51} < 3^{33}$$ which implies the original inequality.

If we are allowed to "prove" with a calculator, then we have

$$\ln 2^{50}=50\ln2\approx 34.65735903$$

and

$$\ln 3^{33}=33\ln 3\approx 36.25420553$$

As $\ln2^{50}<\ln3^{33}$, $2^{50}<3^{33}$.

For the generalization. Note that

$$3^{50}=9^{25}>8^{25}=4^{\frac{3}{2}\times 25}>4^{33}$$

So it is not true for $n=3$.

Note that $n^{50}>(n+1)^{33}$ if and only if $50\ln n>33\ln(n+1)$.

Let $f(x)=50\ln x-33\ln(x+1)$. Then

$$f'(x)=\frac{50}{x}-\frac{33}{x+1}=\frac{17x+50}{x(x+1)}$$

which is positive for all $x>0$. $f$ is strictly increasing for $x>0$.

As $3^{50}>4^{33}$, $f(3)>0$. Hence $f(x)>0$ for all $x\ge 3$.

The inequality $n^{50}<(n+1)^{33}$ does not hold for $n>2$.

$3^7>2^{11}$.

Thus, $3^{35}>2^{55}$ and since $2^5>3^2$, we are done!

• @MartinR Right, I was thinking of the way some of the other answers worked, which didn't have such a step. – David K May 11 '17 at 12:28