The aim of this question is to show this lemma:

Prove that $(√3+2)^{m}$ is not a natural number for all natural numbers $m≥1$.


Because $(\sqrt{3}+2)^{m}+(-\sqrt{3}+2)^{m}$ is an integer and $0<(-\sqrt{3}+2)^{m}<1$.

  • $\begingroup$ Note that this also shows that the expression becomes arbitrarily close to an integer. $\endgroup$ – Mark Bennet Oct 25 '14 at 17:32
  • $\begingroup$ and compute the fractional part of this number... $\endgroup$ – Chen Jiang Oct 25 '14 at 17:44
  • $\begingroup$ @ChenJiang: can you please elaborate about this last comment. $\endgroup$ – DER Nov 5 '14 at 17:00
  • 1
    $\begingroup$ a+b is an integer and 0<b<1 implies that the fractional part of a is just 1-b $\endgroup$ – Chen Jiang Nov 6 '14 at 3:06

Using the Binomial Theorem, we have:

\begin{align*} \sqrt{3}^m + \binom{m}{1}\sqrt{3}^{m-1}2+\cdots+\binom{m}{m-1}\sqrt{3}2^{m_1}+2^m \end{align*}

The even $m$'s will become rational, but the odd $m$'s will make the $\sqrt{3}$ become a radical, just raised to a higher power. They cannot cancel out because it will always be positive.

  • $\begingroup$ This isn't enough, there's no reason to think that the odd powers of $\sqrt 3$ won't add up to an integer. $\endgroup$ – Git Gud Oct 25 '14 at 17:18
  • 1
    $\begingroup$ @GitGud Assume $k$ is odd then $\sqrt{3}^k=\sqrt{3}^{2n+1}=3^n\cdot\sqrt3$. So those odd parts can be written as $q\sqrt3$ for some whole number $q$, and factoring will yield $\sqrt{3}\cdot\text{a whole number}$ which is irrational. $\endgroup$ – Hakim Oct 25 '14 at 17:21
  • $\begingroup$ @Hakim I had missed that, but I stand by my comment, what the answer has written isn't enough. $\endgroup$ – Git Gud Oct 25 '14 at 17:23
  • $\begingroup$ @GitGud Could you clarify why please? Maybe I also missed a point. $\endgroup$ – Hakim Oct 25 '14 at 17:31
  • $\begingroup$ @Hakim I mean that it's not sufficiently justified, specially considering the answer mentions the non-negativity of numbers which implies the impossibility of cancelation. $\endgroup$ – Git Gud Oct 25 '14 at 17:34

I would like to see a comment on if this solution can be improved:

Assume $(\sqrt{3} + 2)^m = p$ so that it is an integer, $n > 0$, a natural number.

$\sqrt{3} + 2 = {p}^{1/m}$

$2 = {p}^{1/m} - \sqrt{3}$

So, in other words, $p$ is a natural number, taken so some power $m \ge 1$, but.

$p^{1/m} - \sqrt{3}$ is irrational, which means, $2$ is irrational. A contradiction.

  • $\begingroup$ the last sentence is not clear. "irrational" part $\endgroup$ – Chen Jiang Oct 25 '14 at 17:30
  • $\begingroup$ @ChenJiang. rational plus irrational is irrational. And if $p^{1/m}$ is irrational! the sum will still be irrational On the LHS, I had $2$ concluding that the RHS is irrational would conclude that the LHS $=2$ is irrational as well! which is a false contradiction. $\endgroup$ – Amad27 Oct 25 '14 at 18:21
  • $\begingroup$ "And if $p^{1/m}$ is irrational! the sum will still be irrational" this sentence is not so clear. $\endgroup$ – Chen Jiang Oct 25 '14 at 18:23
  • $\begingroup$ @ChenJiang,sorry, I'll try to be clear here. If $p^{1/m}$ is irrational, provided that $p^{1/m} \ne \sqrt{3}$ the sum, $p^{1/m} - \sqrt{3} = x$ will prove to be that $x$ is in irrational number, which is of course, not a natural number. This concludes the RHS is irrational. But the LHS = $2$, which means that it says if RHS is irrational then LHS is irrational, which is false because $2$ is rational, a natural number. $\endgroup$ – Amad27 Oct 25 '14 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.