I haven't seen really a straightforward proof towards this question. All of them regarding this topic focus on the fact that $x+y$ can be rational even if x and y are irrational because you could set y as the negative of x, but there isn't really anything about if both are positive and have no minus signs "inside" the variable (say 10-sqrt2 is not valid).
[Math] Prove that $x+y$ is irrational if x and y are irrational and positive.
irrational-numbers
Related Solutions
This proof is from The Transcendence of $\pi$ by Steve Mayer, November 2006
I've rewritten it below partly so that I understand it and partly so that if you're too lazy to click on the link or if the link goes away there's a copy here.
Definition: A complex number is algebraic over $\mathbb{Q}$ if it is a root of a polynomial equation with rational coefficients.
i.e: $a$ is algebraic if there are rational numbers $\alpha_0, \alpha_1,...,\alpha_n$ not all 0 such that $\alpha_0a^n + \alpha_1a^{n-1} + ... + \alpha_{n-1}a + \alpha_n = 0$
Definition: A complex number is transcendental if it is not algebraic.
Theorem (Lindemann-Weierstrass): $\pi$ is transcendental over $\mathbb{Q}$
Proof: If $\pi$ satisfies an algebraic equation with coefficients in $\mathbb{Q}$, so does $i\pi$. Let this equation be $\theta_1(x) = 0$ with roots $i\pi = \alpha_1, ..., \alpha_n.$ Now, $e^{i\pi} + 1 = 0$ so
$(e^{\alpha_1} + 1)...(e^{\alpha_n} + 1) = 0$
We now construct an algebraic equation with integer coefficients whose roots are the exponents of $e$ in the expansion of the above product. For example, the exponents in pairs are $\alpha_1 + \alpha_2, \alpha_1 + \alpha_3, ..., \alpha_{n-1} + \alpha_n$. The $\alpha$'s satisfy a polynomial equation over $\mathbb{Q}$ so their elementary symmetrix functions are rational. Hence the elementary symmetric functions of the sums of pairs are symmetric functions of the $\alpha$'s and are also rational. Thus the pairs are the roots of the equation $\theta_2(x) = 0$ with rational coefficients. Similarly, sums of 3 $\alpha$'s are roots of $\alpha_3(x) = 0$, etc...
Then, the equation
$\theta_1(x)\theta_2(x)...\theta_n(x) = 0$
is a polynomial equation over $\mathbb{Q}$ whose roots are all sums of $\alpha$'s. Deleting zero roots from this, if any, we get:
$\theta(x) = 0$
$\theta(x) = cx^r + c_1x^{r-1}+...c_r$
and $c_r \neq 0$ since we have deleted zero roots. The roots of this equation are the non-zero exponents of $e$ in the product when expanded. Call these $\beta_1,...,\beta_r$. The original equation becomes
$e^{\beta_1} + ... e^{\beta_r} + e^0 + ... e^0 = 0$
i.e: $\sum e^{\beta_i} + k = 0$
where k is an integer $> 0$ ($\neq 0$ since the term $1...1$ exists)
Now define
$f(x) = c^sx^{p-1}\frac{[\theta(x)]^p}{(p-1)!}$
where $s = rp-1$ and $p$ will be determined later.
Define:
$F(x) = f(x) + f'(x) +...+f^{(s+p)}(x)$
Then,
$\frac{d}{dx}[e^{-x}F(x)] = - e^{-x}f(x)$
Hence, we have
$e^{-x}F(x) - F(0) = - \int\limits_{0}^{x}e^{-y}f(y)dy$
Putting $y = \lambda x$ we get
$F(x) - e^xF(0) = -x \int\limits_{0}^{1}e^{(1-\lambda)x}f(\lambda x)d\lambda$
Let $x$ range over the $\beta_i$ and sum. Since $\sum e^{\beta_i} + k = 0$ we get
$\sum\limits_{j = 1}^{r} F(\beta_j) + kF(0) = - \sum\limits_{j=1}^{r} \beta_j \int\limits_{0}^{1} e^{(1-\lambda)\beta_j} f(\lambda \beta_j)d\lambda$
CLAIM For large enough $p$ the left hand size is a non-zero integer.
$\sum\limits_{j=1}^{r} f^{(t)}(\beta_j) = 0$ $(0 < t < p)$ by definition of $f$. Each derivative of order p or more has a factor $p$ and a factor $c^s$, since we must differentiate $[\theta(x)]^p$ enough times to not get $0$. And $f^{(t)}(\beta_j)$ is a polynomial in $\beta_j$ of degree at most $s$. The sume is symmetric, and so in an integer provided each coefficient is divisible by $c^s$, which it is. (symmetric functions are polynomials in coefficients $=$ polynomials in $\frac{c_i}{c}$ of degree $\leq s$). Thus we have,
$\sum\limits_{j=1}^r f^{(t)}(\beta_j) = pk_t$
$t = p,...,p+s$
Thus, the left hand side $LHS=$ (integer) = $kF(0).$ What is $F(0)?$
$f^{(t)}(0) = 0$ $t = 0, ...,p-2$
$f^{(p-1)}(0) = c^sc_r^p$ $(c_r \neq 0)$
$f^{(t)}(0) = p$(some integer) $t = p, p+1,...$
So, $LHS$ is an integer multiple of $p+c^sc_r^pk$. This is not divisible by $p$ if $p > k, c, c_r$. So it is a non-zero integer. But the right hand side tends to $0$ as $p \rightarrow \infty$ and thus we get a contradiction and thus $\pi$ is transcendental.
You can use the convergents to the continued fraction to obtain series, since the convergents give close rational approximations to $5$. $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$
Using $\sqrt{5} \approx \lfrac{9}{4}$ we get:
$\sqrt{5} = \lfrac94 ( 1 - \lfrac1{9^2} )^\lfrac12 = \lfrac{9}{4} \sum_{k=0}^\infty \binom{1/2}{k} \lfrac{(-1)^k}{9^{2k}} = \lfrac{9}{4} - \lfrac{9}{4} \sum_{k=1}^\infty \lfrac{2 \cdot (2k-2)!}{k! \cdot (k-1)! \cdot 18^{2k}}$.
Using $\sqrt{5} \approx \frac{38}{17}$ we get:
$\sqrt{5} = \lfrac{38}{17} ( 1 + \lfrac1{38^2} )^\lfrac12 = \lfrac{38}{17} \sum_{k=0}^\infty \binom{1/2}{k} \lfrac{1}{38^{2k}} = \lfrac{38}{17} - \lfrac{38}{17} \sum_{k=1}^\infty \lfrac{(-1)^k \cdot 2 \cdot (2k-2)!}{k! \cdot (k-1)! \cdot 76^{2k}}$.
Best Answer
You may take $\sqrt 2$ and $10-\sqrt 2$, they are both positive and irrational and their sum is $10$