1
$\begingroup$

Prove that $2\sqrt2 + \sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2\sqrt2 + \sqrt7 = \frac{1}{2\sqrt2 - \sqrt7}$.

Assume $2\sqrt2 + \sqrt7 $ is rational. Then

$2\sqrt2 + \sqrt7 = \frac{m}{n}$

and

$2\sqrt2 - \sqrt7 = \frac{n}{m}$

if I add them together I get

$ 4\sqrt2 = \frac{m^2+n^2}{mn} $

then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $\sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?

$\endgroup$
0
$\begingroup$

The proof works because if your number is a rational $\frac mn$, then its reciprocal, $\frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2\sqrt2 = -\sqrt7$, which is clearly false.)

Similarly, when you add $\frac mn$ to $\frac nm$, you get another rational number, $\frac{m^2+n^2}{mn}$.

$\endgroup$
  • $\begingroup$ In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes. $\endgroup$ – Théophile Oct 17 '15 at 22:26
0
$\begingroup$

It is pure luck that $2\sqrt2 + \sqrt7 = \frac{1}{2\sqrt2 - \sqrt7}$, down to the fact that $(2\sqrt 2)^2-(\sqrt 7)^2 = 1$.

The following approach is more genral:

Suppose $2\sqrt2 + \sqrt7$ is rational. Then its square, $15+4\sqrt {14}$, is still rational. So $\sqrt{14}$ must be rational. Which it isn't.

$\endgroup$
0
$\begingroup$

Here's another proof more general that doesn't rely on friendly symmetry. $\sqrt 2$ and $\sqrt 7$ are easy to show to be irrational. So for $2\sqrt 2 + \sqrt 7 = r$ then $2\sqrt 2 - r = -\sqrt 2$ so $8 + r^2 + 2\sqrt 2 = 7$ therefore $sqrt 2$ is rational.

It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.

$\endgroup$
  • $\begingroup$ I have noticed the following mistakes in your answer. $2\sqrt(2) - r$ is $-\sqrt(7)$ and not $-\sqrt(2)$. Also $8+r^2 + 2\sqrt(2)$ is $23+2\sqrt(2)+4\sqrt(14)$ and not $7$. $\endgroup$ – Bysshed Apr 2 '16 at 22:58
-1
$\begingroup$

From

$$2\sqrt2 + \sqrt7 = \frac{n}{m}$$ and $$2\sqrt2 - \sqrt7 = \frac{m}{n},$$ by addition, you get

$$4\sqrt{2}=\frac{m}{n}+\frac{n}{m}.$$ That is, $\sqrt{2}$ should be rational.

${}{}{}{}$

$\endgroup$
  • $\begingroup$ I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part. $\endgroup$ – Miguelgondu Oct 17 '15 at 23:39
  • $\begingroup$ @Miguelgondu Typo corrected. Thank you for noticing it. $\endgroup$ – mfl Oct 18 '15 at 4:01

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.