Do there exist real numbers $a, b$ $a,b$ such that (1) $a + b$ $a+b$ is rational and $a^{n} + b^{n}$ $a^n+b^n$ is irrational for each natural $n \geq 2$ $n ge 2$ (2) $a + b$ $a+b$ is irrational and $a^{n} + b^{n}$ $a^n+b^n$ is rational for each natural $n \geq 2$ $n ge 2$ ?

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Do there exist real numbers $a, b$ $a,b$ such that (1) $a + b$ $a+b$ is rational and $a^{n} + b^{n}$ $a^n+b^n$ is irrational for each natural $n \geq 2$ $n ge 2$ (2) $a + b$ $a+b$ is irrational and $a^{n} + b^{n}$ $a^n+b^n$ is rational for each natural $n \geq 2$ $n ge 2$ ?

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Answer 1 · 2017-09-11T21:12:32

(1) Yes, e.g. $a = π$ $a=pi$ , $b = 1 - π$ $b=1-pi$

(2) No, see explanation.

Explanation:

(1) Let $a = π$ $a=pi$ , $b = 1 - π$ $b=1-pi$

Then:

$a + b = 1$ $a+b = 1$ is rational

When $n \geq 2$ $n >= 2$

$a^{n} + b^{n} = π^{n} + {(1 - π)}^{n}$ $a^n+b^n = pi^n+(1-pi)^n$

is a non-trivial polynomial in $π$ $pi$ , hence irrational, since $π$ $pi$ is transcendental.

$color(white)()$
(2) Suppose $a^{n} + b^{n}$ $a^n+b^n$ is rational for all $n \geq 2$ $n >= 2$

If $a = b = 0$ $a=b=0$ then $a + b$ $a+b$ is rational.

If $a = 0$ $a=0$ and $b \neq 0$ $b!=0$ or $a \neq 0$ $a!=0$ and $b = 0$ $b=0$ then:

$\frac{a^{3} + b^{3}}{a^{2} + b^{2}} = a + b$ $(a^3+b^3)/(a^2+b^2) = a+b" "$ is rational

Otherwise suppose $a, b \neq 0$ $a, b != 0$

Then:

${(a^{2} + b^{2})}^{2} - (a^{4} + b^{4}) = 2 a^{2} b^{2}$ $(a^2+b^2)^2-(a^4+b^4) = 2a^2b^2$

The left hand side is rational, so the right hand side must be.

Hence $a^{2} b^{2}$ $a^2b^2$ is rational.

Then:

$(a^{2} + b^{2}) (a^{3} + b^{3}) - (a^{5} + b^{5}) = a^{2} b^{2} (a + b)$ $(a^2+b^2)(a^3+b^3)-(a^5+b^5) = a^2b^2(a+b)$

The left hand side of this equation is rational and the multiplier $a^{2} b^{2}$ $a^2b^2$ is also rational (and non-zero). So $a + b$ $a+b$ is rational.

Answer 2 · 2017-09-12T12:56:38

(1) Let $a = \sqrt{2}$ $a=sqrt(2)$ and $b = 1 - \sqrt{2}$ $b=1-sqrt(2)$

Explanation:

Here's an alternative answer for (1) using simple irrational numbers:

Let $a = \sqrt{2}$ $a = sqrt(2)$ , $b = 1 - \sqrt{2}$ $b=1-sqrt(2)$

Then:

$a + b = 1$ $a+b = 1" "$ is rational

Note that:

$(p + q \sqrt{2}) (1 - \sqrt{2}) = (p - 2 q) + (q - p) \sqrt{2}$ $(p+qsqrt(2))(1-sqrt(2)) = (p-2q)+(q-p)sqrt(2)$

So consider the recursively defined sequences:

$p_{0} = 1$ $p_0 = 1$

$q_{0} = 0$ $q_0 = 0$

$p_{n + 1} = p_{n} - 2 q_{n}$ $p_(n+1) = p_n-2q_n$

$q_{n + 1} = q_{n} - p_{n}$ $q_(n+1) = q_n-p_n$

Then:

$p_{n} + q_{n} \sqrt{2} = {(1 - \sqrt{2})}^{n}$ $p_n + q_nsqrt(2) = (1-sqrt(2))^n$ for $n \geq 0$ $n >= 0$
The sequences begin:

$1, 1, 3, 7, 25, 57, 139,...$

$0, -1, -2, -9, -16, -41, -98,...$

$p_n > 0$ for all $n >= 0$ and is always odd.
$q_n < 0$ for all $n >= 1$ and alternates between odd and even values.

We can also define sequences for the $a^n$ term, by:

$r_0 = 1$

$s_0 = 0$

$r_(n+1) = 2s_n$

$s_(n+1) = r_n$

Then:

$r_n + s_nsqrt(2) = sqrt(2)^n$ for $n >= 0$
These sequences begin:

$1, 0, 2, 0, 4, 0, 8,...$

$0, 1, 0, 2, 0, 4, 0,...$

Note that when $n >= 2$ , both $r_n$ and $s_n$ are even.

When $n$ is even then $s_n = 0$ , so $q_n+s_n = q_n < 0$ for $n >= 2$ .

When $n$ is odd then $q_n$ is odd so $q_n+s_n$ is odd and non-zero.

Hence in all cases, when $n >= 2$ we find:

$sqrt(2)^n + (1-sqrt(2))^n = (p_n+r_n) + (q_n+s_n)sqrt(2)$

with $q_n+s_n != 0$ and hence irrational.

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Do there exist real numbers $a, b$ $a,b$ such that (1) $a + b$ $a+b$ is rational and $a^{n} + b^{n}$ $a^n+b^n$ is irrational for each natural $n \geq 2$ $n ge 2$ (2) $a + b$ $a+b$ is irrational and $a^{n} + b^{n}$ $a^n+b^n$ is rational for each natural $n \geq 2$ $n ge 2$ ?

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