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Question:

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Following from the question, I tried:

Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,

$$ \begin{align} N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\ &= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\ &= (1+\sqrt{3})(1+\sqrt{7}). \end{align} $$

Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.

But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.

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    $\begingroup$ Duplicate of Irrationality of $(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$ $\ \ $ $\endgroup$ May 7, 2019 at 15:37
  • $\begingroup$ For generalizations I recommend this. Bill Dubuque's answer in particular uses the minimum amount of theory. $\endgroup$ May 7, 2019 at 15:42
  • $\begingroup$ @Jyrki More precisely, my answer translates (Kummer) Galois theory into more elementary language. At OP: if you study abstract algebra and/or number theory then you will learn that results like this are essentially special cases of Galois theory of Kummer extensions. But one needs only basic knowledge of fields to understand my answer (and even that could be eliminated to get a high-school level proof). $\endgroup$ May 7, 2019 at 15:49

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If $(1+\sqrt{3})(1+\sqrt{7})$ is rational, then

$$\displaystyle \frac{12}{(1+\sqrt{3})(1+\sqrt{7})}=\frac{12(1-\sqrt{3})(1-\sqrt{7})}{(-2)(-6)}=1-\sqrt{3}-\sqrt{7}+\sqrt{21}$$ is also rational.

So, $\displaystyle \frac{1}{2}[(1+\sqrt{3})(1+\sqrt{7})+1-\sqrt{3}-\sqrt{7}+\sqrt{21}]-1=\sqrt{21}$ is rational.

This leads to a contradiction.

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Suppose $(1+\sqrt3)(1+\sqrt7)=p/q$ for some $p,q\in\Bbb Z^+$. Then we have that $$q(1+\sqrt3)=\frac p{1+\sqrt7}=\frac{p(1-\sqrt7)}{-6}\implies p\sqrt7-6q\sqrt3=p+6q\ne0\tag1$$ This implies that $$p\sqrt7+6q\sqrt3=\frac{(p\sqrt7+6q\sqrt3)(p\sqrt7-6q\sqrt3)}{p\sqrt7-6q\sqrt3}=\frac{7p^2-108q^2}{p+6q}\tag2$$ Adding $(1)$ and $(2)$ together gives $$2p\sqrt7=p+6q+\frac{7p^2-108q^2}{p+6q}\implies\sqrt7\in\Bbb Q$$ which is a contradiction. $\square$

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  • $\begingroup$ One should remark / prove that the denominators are nonzero (this is where such proofs may fail if the square roots are not linearly independent over $\Bbb Q)\ \ $. $\endgroup$ May 8, 2019 at 22:36
  • $\begingroup$ @BillDubuque In this case it should be obvious since $p\sqrt7-6q\sqrt3=p+6q>0$. $\endgroup$
    – TheSimpliFire
    May 9, 2019 at 6:43
  • $\begingroup$ Yes, for these particular radicands and coef's it is very easy. But for others it may be far less obvious. It deserves attention since it plays a crucial role in the correctness of the proof (and there is often oversight about such by beginners, i.e. I'm not simply nitpicking). $\endgroup$ May 9, 2019 at 13:26
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$$\begin{eqnarray*} N=\sqrt{3}+\sqrt{7}+\sqrt{21} & \Rightarrow & N-\sqrt{21}=\sqrt{3}+\sqrt{7} \\ & \Rightarrow & {{N}^{2}}+21-2N\sqrt{21}=10+2\sqrt{21} \\ & \Rightarrow& \sqrt{21}=\frac{{{N}^{2}}+11}{2+2N} \\ \end{eqnarray*}$$

So $\sqrt{21}$ is rational, which is a contradiction

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A somewhat systematic (but laborious) approach: Assume $$N=\sqrt 3+\sqrt 7+\sqrt{21} $$ is rational. Then also $$N^2=3+7+21+2(\sqrt{21}+3\sqrt 7+7\sqrt 3)= 31+2\sqrt{21}+3\sqrt 7+7\sqrt 3$$ is rational, as well as $$(N^2-31)^2 =4\cdot 21+9\cdot 7+49\cdot 3+2(42\sqrt 3+42\sqrt 7+21\sqrt{21}).$$ Thus also $$(N^2-31)^2- (4\cdot 21+9\cdot 7+49\cdot 3)-84N=-42\sqrt{21}$$ is rational. I guess you can see how this could be similarly applied to all specific sums of square roots ...

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  • $\begingroup$ seems like there are some mistakes in calculation, i am getting confused, for this one $\endgroup$ May 7, 2019 at 15:47
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In this answer I proved the theorem below about positive sums of square roots in ordered fields.

Theorem $\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} \in K\ \Rightarrow \sqrt{c_i}\in K\,$ for all $\, i,\:$ if $\,0 < c_i\in K$ an ordered field.

It is instructive to specialize that proof here. Readers unfamiliar with fields should begin here.

Assuming $\, q = \sqrt 3 +\!\sqrt 7+\!\sqrt{21} \in\Bbb Q\,$ we infer $\color{#0a0}{\text{one of }\,\sqrt3,\sqrt 7,\sqrt{21}\,\ {\rm is\ in}\ \Bbb Q},\,$ a contradiction.

By the Lemma below $\, \sqrt{7} + \sqrt{21} = q\!-\!\sqrt 3 \in \Bbb Q(\sqrt 3)\,\Rightarrow\, \sqrt{7}, \sqrt{21} \in \Bbb Q(\sqrt 3),\,$ thus

$$\begin{align} \sqrt{7}\, &= \,a\, +\, b\sqrt 3,\ \ \ \ a,\,b\ \in\Bbb Q\\[.3em] \sqrt{21}\, &=\, a' + b'\sqrt 3,\ \ \ a',b'\in\Bbb Q\end{align}\qquad\quad\ \ \ $$

If $\, b\, < 0\,$ then $\,a = \sqrt 7 - b\sqrt 3 = \sqrt 7 + \sqrt{3b^2}\in\Bbb Q\Rightarrow\color{#0a0}{\sqrt 7\in \Bbb Q}\,$ by the Lemma.
If $\ b'\! < 0\,$ then the same argument allows us to deduce $\color{#0a0}{\sqrt{21}\in \Bbb Q}$
Else $\,b,b'\ge 0\,\Rightarrow\,\color{#c00}{1\!+\!b\!+\!b' > 0}\,$ so, by below, we infer $\, \color{#0a0}{\sqrt 3\,\in \Bbb Q},\,$ a $\rm\color{#0a0}{\rm contradiction}$ in all cases. $\quad\ q = \sqrt 3 +\!\sqrt 7+\!\sqrt{21} = a\!+\!a'+(1\!+\!b\!+\!b')\sqrt 3\,\ $ so $\ \sqrt 3 = \dfrac{q\!-\!a\!-\!a'}{\color{#c00}{1\!+\!b\!+\!b'}}\in\Bbb Q.\ \ \bf\small QED$


Lemma $ $ If $\,0< r,s\in K$ then $\, k = \sqrt r +\!\sqrt s\in K\Rightarrow\sqrt r,\sqrt s\in K,\,$ for any subfield $\,K\subseteq \Bbb R$

Proof $\ $ Note $\ k' = \sqrt{r}-\!\sqrt{s}\: = \dfrac{\ \ r\, -\ s}{\sqrt{r}+\!\sqrt{s}}\in K,\,$ by $\,0 < \sqrt r +\! \sqrt s\in K,\,$ by $\, \sqrt r,\sqrt s > 0$

Therefore $\ (k+k')/2 = \sqrt r\in K\ $ and $\ (k-k')/2 = \sqrt s\in K$.

Remark $ $ Above is the inductive step of the above linked general proof specialized to the case $\,P(2)\Rightarrow P(3),\,$ where $\,P(n)\,$ denotes the proposition with a sum of $\,n\,$ square roots. The general induction step works precisely the same way.

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$\sqrt{21}+\!\sqrt{7}+\!\sqrt{3}=q\,\Rightarrow\, \sqrt7(\sqrt 3+1) = q-\sqrt3 \,\Rightarrow\, \sqrt 7\in \Bbb Q(\sqrt3),\,$ contra Lemma below.


Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $ all are not in $\rm\,K\,$ and $\rm\, 2 \ne 0\,$ in the field $\rm\,K.$

Proof $\ $ Let $\rm\ L = K(\sqrt{b}).\,$ $\rm\, [L:K] = 2\,$ via $\rm\,\sqrt{b} \not\in K,\,$ so it suffices to prove $\rm\, [L(\sqrt{a}):L] = 2.\,$ It fails only if $\rm\,\sqrt{a} \in L = K(\sqrt{b})\, $ and then $\rm\, \sqrt{a}\ =\ r + s\, \sqrt{b}\ $ for $\rm\ r,s\in K.\,$ But that's impossible,
since squaring $\Rightarrow \rm\color{#c00}{[\![1]\!]}\!:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\ \sqrt{b},\, $ which contradicts hypotheses as follows:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $\color{#c00}{[\![1]\!]}$ for $\rm\sqrt{b},\,$ using $\rm\,2 \ne 0$

$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r+s\,\sqrt b = r \in K$

$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\,b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\, \sqrt{b},\ \ $times $\rm\,\sqrt{b}\quad$

Remark $ $ This method generalizes to sums of any number of sqrts. See the citations there for generalizations to $n$'th roots. Readers unfamilier with field theory may consult this answer.

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