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.

Question

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.

Answer 1

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.

Answer 2

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. $\square$

Answer 3

$$\begin{eqnarray*} N=\sqrt{3}+\sqrt{7}+\sqrt{21} & \Rightarrow & N-\sqrt{21}=\sqrt{3}+\sqrt{7} \\ & \Rightarrow & {{N}^{2}}+21-2N\sqrt{21}=10+2\sqrt{21} \\ & \Rightarrow& \sqrt{21}=\frac{{{N}^{2}}+11}{2+2N} \\ \end{eqnarray*}$$

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

Answer 4

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 ...

Answer 5

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

Answer 6

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.