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  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.

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    $\begingroup$ 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. $\endgroup$ – user May 18 at 8:51
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    $\begingroup$ Technically, $31/13$ is closer to $2$ than it is to $3$. $\endgroup$ – Saucy O'Path May 18 at 8:51
  • $\begingroup$ @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 $\endgroup$ – user477465 May 18 at 8:58
  • $\begingroup$ It is clear that $\sqrt{13}$ is not integer as $3<\sqrt{13}<4$. Therefore it is irrational. $\endgroup$ – user May 18 at 9:03
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    $\begingroup$ 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. $\endgroup$ – TonyK May 18 at 9:06
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$\sqrt{13}$ 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 $2^{24}<10^8<5^{12}$.

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  • $\begingroup$ 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? $\endgroup$ – user130306 May 18 at 20:27
  • $\begingroup$ $\frac{10^8}{5^{12}}<1$ implies $10^8<5^{12}$ and $\frac{10^8}{2^{24}}>1$ implies $2^{24}<10^8$. $\endgroup$ – Hongyi Huang May 19 at 2:45
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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 $5^{12} \gt 10^8 \gt 2^{24}$

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$$\sqrt{13}$$ 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.

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