Prove by contradiction that $(a + b + 1) ^ {\frac {1} {a + b}} $ is irrational

Question

Suppose if it is rational:

$9 ^ {\frac {1} {8}} = {\frac {m} {n}}$

I know what to do with relative primes. M and N are the relative primes.

$(n \times 9 ^ {\frac {1} {8}}) = m $

$(n \times 9 ^ {\frac {1} {8}}) ^ 8= m ^ 8 $

$n ^ 8 \times 9= m ^ 8 $

$n ^ 8 \times 3 ^2= m ^ 8$

I assume that M will be bigger than N. For example, $m ^ 8 = 3^8 \times 5^8 \times 7^8. $

I know it will make a contradiction so therefore it is irrational. My question to myself is "does it makes sense to see a proof that it is not rational?"

I just want to make clear contradiction but it is hard for me to explain. Can you help me? Thank you!

Answer 1

For bigger $a, b$ than in @Doug's counterexample, write: $(a+b+1) = (\frac{m}{n})^{a+b}.$

Rearranging terms, get,

$$m^{a+b} = n^{a+b}(a+b+1).$$

Since $m, n$ are relatively prime, any prime divisor of $a+b+1$ has to appear to $a+b$-th power. The smallest prime divisor possible is $2,$ but $2^{a+b} > a+b+1,$ UNLESS $a+b=1.$

Answer 2

It's not true. Let a=0 and b=1. Let b=0 and a=1. Both are rational.

Answer 3

Suppose it is rational an put $x = a + b + 1$, then $a + b = x - 1 $. wlog assume $a + b > 1 $

$$ \therefore x^{\frac{1}{1-x}} = \frac{m}{n} $$

$$ x = (\frac{m}{n})^{x-1} = \frac{n}{m}(\frac{m}{n})^x $$

Claim: A line $x$ never intersects with the function $f(x) = \frac{1}{r} r^x $ where $r \in \mathbb{Q}^{>0} $ if $x > 1 $.

Indeed if they match, then

$$ x = \frac{1}{r} r^x \implies rx = r^x \implies \ln rx = x \ln r \implies \frac{ \ln rx }{rx} = \frac{ \ln r }{ r}$$. Notice the only solution to this is $x = 1$, and $x > 1 $ never solves this equation.

Now, apply claim to your problem to see that

$$ x = (\frac{m}{n})^{x-1} = \frac{n}{m}(\frac{m}{n})^x $$

is impossible.