# How can I order these numbers without a calculator?

1. Classify the following numbers as rational or irrational. Then place them in order on a number line:

$$\pi^2, -\pi^3, 10, 31/13, \sqrt{13}, 2018/2019, -17, 41000$$

I know $$\pi$$ is irrational so $$\pi^2$$ and $$-\pi^3$$ are irrational. 10, 31/30, 2018/2019, -17, 41000 are all rational because then can be written in the form $$\frac{a}{b}, a,b \in \mathbb{Z}, b\neq 0$$

I'm not sure how I can look at $$\sqrt{13}$$ and determine if it's rational or irrational.

For ordering them I am approximating $$\pi$$ as 3, so

$$\pi^2 \approx 9$$ and

$$-\pi^3 \approx -27$$.

$$31/13 \approx 30/10 = 3$$

$$\sqrt{9}<\sqrt{13}<\sqrt{16}$$ so $$3<\sqrt{13}<4$$

$$2018/2019 \approx 1$$

If I were to order these, I would say:

$$-\pi^3, -17, 2018/2019, 31/13, \sqrt{13}, \pi^2, 41000$$

1. Put these numbers in order from least to greatest (no calculator):

$$10^8, 5^{12}, 2^{24}$$

All I can think of is that $$2^{10} \approx 10^3$$ so

$$2^{24} = (2^{10})^{2}* 2^{4} \approx (10^3)^2 *2^4 = 10^6 *16 <10^{8}$$

So $$2^{24} < 10^8$$

Not sure how $$5^{12}$$ fits in.

• From $\pi$ is irrational does not follow that $\pi^2$ is irrational, think about $\sqrt2$. $\pi$ is however transcendental, so every its power is transcendental (and therefore irrational). $\sqrt{13}$ is irrational since it is not integer. – user May 18 at 8:51
• Technically, $31/13$ is closer to $2$ than it is to $3$. – Saucy O'Path May 18 at 8:51
• @user ah thank you for clarification of the $\pi^2$ stuff. However for $\sqrt{13}$ i don't have a calculator to evaluate, is there any other way to see that it's irrational? and for $31/13$ it would probably be better for me to approximate it to $30/14$ and then reduce it – user477465 May 18 at 8:58
• It is clear that $\sqrt{13}$ is not integer as $3<\sqrt{13}<4$. Therefore it is irrational. – user May 18 at 9:03
• Did you know that if $n$ is not a perfect square, then $\sqrt n$ is irrational? You can find a proof in this Wikipedia article. – TonyK May 18 at 9:06

$$\sqrt{13}$$ is irrational, by user.

The order of your first question should be $$-\pi^3,-17,2018/2019,31/13,\pi^2,10,41000$$.

$$-\pi^3<-3^3 = -27<-17$$, so we can compare two negative numbers.

We now compare $$\pi^2$$ and $$10$$, others are easy to compare. Note that $$\pi = 3.1415926\dots$$ and so $$\pi<3.15$$. Now $$3.15^2 = 9.9225<10$$, so $$\pi^2<3.15^2<10$$.

For your second question, just divide them respectively.

$$\frac{10^8}{5^{12}} = \frac{2^85^8}{5^{12}} = \frac{2^8}{5^4} = \frac{4^4}{5^4}<1$$, and $$\frac{10^8}{2^{24}} = \frac{2^85^8}{2^{24}} = \frac{5^8}{2^{16}} = \frac{5^8}{4^8}>1$$. Thus $$2^{24}<10^8<5^{12}$$.

• I understand why $2^{24} < 5^{12}$ from your solution but I'm confused about how I kow that $10^8$ is in the middle? – user130306 May 18 at 20:27
• $\frac{10^8}{5^{12}}<1$ implies $10^8<5^{12}$ and $\frac{10^8}{2^{24}}>1$ implies $2^{24}<10^8$. – Hongyi Huang May 19 at 2:45

For the second list we can see that $$5 \gt 2^2$$ as $$5 \gt 4$$ so if we take both sides to the power of $$12$$ we get $$5^{12} \gt 2^{24}$$. Then we can compare $$10^8$$ and $$2^{24}$$, we can split $$10$$ into $$2*5$$, so $$10^8=2^8*5^8$$. We claim $$10^8 \gt 2^{24}$$, which leads to $$2^8*5^8 \gt 2^{24}$$, then taking $$2^8$$ over, $$5^8 \gt 2^{16}$$. Then by our logic before $$5 \gt 2^2$$ we can prove the claim!

So the order is $$5^{12} \gt 10^8 \gt 2^{24}$$

$$\sqrt{13}$$ cannot be rational. Because if it was, you could write

$$\sqrt{13}=\dfrac pq$$ where $$p,q$$ are not both multiples of $$13$$ (otherwise you could simplify). From this

$$p^2=13q^2.$$

This implies that $$p^2$$ is a multiple of $$13$$, and so must $$p$$ be, as $$13$$ is a prime. Then $$p^2$$ is a multiple of $$169$$, so that $$q^2$$ is a multiple of $$13$$. And so is $$q$$, a contradiction.