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I took $\sqrt{3}+\sqrt{7}$ and squared it. This resulted in a new value of $10+2\sqrt{21}$.

Now, we can say that $10$ is rational because we can divide it with $1$ and as for $2\sqrt{21}$, we divide by $2$ and get $\sqrt{21}$. How do I prove $\sqrt{21}$ to be rational/irrational?

Thanks

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marked as duplicate by Marc van Leeuwen, Lord_Farin, Daniel Fischer, Nicholas R. Peterson, amWhy Nov 4 '13 at 14:39

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1  
...you timed it by $^2$? What is that supposed to mean? –  Doorknob Nov 4 '13 at 13:16
    
Hope this and this will also help you. –  Sush Nov 4 '13 at 13:24
    
It would be nice if you indicated the strategy you were using. Right now it just looks like you're randomly applying operations to $\sqrt{3}+\sqrt{7}$. For example, you could do this: "Suppose for now that $\sqrt{3}+\sqrt{7}$ is rational. Then it's square is rational. Subtracting 10 from the square is rational, and dividing the result by $2$ is rational. This says that $\sqrt{21}$ is rational. But now we have a contradiction because I can show $\sqrt{21}$ is irrational..." Hope this helps: improving your writing helps both you figure stuff out and the reader figure stuff out. –  rschwieb Nov 4 '13 at 14:14

3 Answers 3

Suppose $\sqrt{7}+\sqrt{3}={a\over b}$ rational number say

then ${4b\over a}={4\over \sqrt{7}+\sqrt{3}}=4{\sqrt{7}-\sqrt{3}\over (\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=(\sqrt{7}-\sqrt{3})$

so $\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}={a\over b}+{4b\over a}$

so $\sqrt{7}={a\over 2b}+{2b\over a}$

so we are getting $\text{ an irrational number }\sqrt{7}=\text{ some rational number}(\Rightarrow\Leftarrow)$

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But if one knows that $\sqrt 7$ is irrational, why does that one not know that $\sqrt {21}$ (as occuring in the question) is also irrational - and omit any further fiddling? I think, in that question is a proof from the scratch required... –  Gottfried Helms Nov 4 '13 at 13:59

Suppose that $$\sqrt 7 +\sqrt 3=r, r\in\mathbb{Q}$$

Now we have: $$\sqrt 7 =r-\sqrt 3/^2$$ $$7=r^2-2\sqrt 3 +3$$ $$4=r^2-2\sqrt 3$$ $$2\sqrt 3=r^2-4$$ $$\sqrt 3=\frac{r^2-4}{2}$$

This is a contradiction, because the left both is $\sqrt 3\in\mathbb{I}$, and the right both is $\frac{r^2-4}{2}\in\mathbb{Q}$. The contradiction is due to make the wrong assumption i.e. $\sqrt 7 +\sqrt 3$ is irational number

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Is $\mathbb{I}$ standard notation for Irrational Number? –  Une Femme Douce Nov 4 '13 at 13:30
    
yes, $\mathbb{I}$ notation for set of irational number –  Madrit Zhaku Nov 4 '13 at 13:31
    
where did you find that? could you tell me? –  Une Femme Douce Nov 4 '13 at 13:32
    
@Sade asks about standard notation. I never saw this in my analysis course. –  Sush Nov 4 '13 at 13:33
    
@Sush Yes, I have never seen that notation for irrationals, I saw $\mathbb{Q}^c$ –  Une Femme Douce Nov 4 '13 at 13:34

@Doorknob: I think he means that he took the square: $(\sqrt{3} + \sqrt{7})^2 = 10 + 2\sqrt{21}$.

If you want to prove that $\sqrt{21}$ is rational/irrational, do as follows. Suppose $a,b \in \mathbb{N}$ are such that $\frac{a}{b} = \sqrt{21}$ and that $\gcd(|a|,|b|) = 1$. I.e., $\frac{a}{b}$ can not be simplified. Then, we must have that $\frac{a^2}{b^2} = 21$. Or: $a^2 = b^2\cdot 21$.

Every prime factor of both $a^2$ and $b^2$ must occur an even amount of times in their respective factorisation. We see, however, that in this case, it will not happen, due to the extra factors 3 and 7 on the right hand side of the equation. This yields a contradiction, thus $\sqrt{21}$ is irrational.

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