1
$\begingroup$

I answered this question but am not quite sure if i did what was correct. If anything is wrong please point it out, thanks.

Question:Prove that any number of the form $a + \sqrt[b]5$ is irrational, where $a,b$ are integers and $b \ne 0$

My Answer:

Proof by Contradiction:

Assume $a + \sqrt[b]5$ is rational:

Assume $p$/$q$ is in lowest terms where GCD($p$,$q$) = $1$ where $p$, $q$ are integers and $q \ne $0

$$\begin{align} a + \sqrt[b]{5} &= {p \over q} & \\\ \sqrt[b]{5} &= \frac pq - a \\ 5 &= \left(\frac pq - a\right)^b \\ 5 &= \frac{(p - qa)^b}{q^b} & \text{(Multiplying q to be the denominator of the whole equation)} \\ 5q^b &= (p - qa)^b &\text{(Isolating q and this shows }(p-qa)^b\text{ is a multiple of 5)} \end{align}$$

Assuming $5c = p - qa$:

$$5q^b = 5^bc^b$$

How do i show q^b is a multiple of 5? Because if the question had a integer other than b then I could prove for odd and even numbers using (5p+1)^b and (5p+2)^b where b is an integer such as 1, 2, ... b then I can expand it and show there is a common divisor of 5 but I cannot do that in this case because I dont know what b is. The assumption to show this equation is rational if and only if b$\sqrt5$ is rational will work but I already knew that my problem is actually showing b$\sqrt5$ is rational.

Then,

$$5q^b = 5\left(5^{b-1}c^b\right) ~~~~\text{(which implies q is a multiple of 5)}$$

Since $q$ is a multiple of $5$:

$$\begin{align} q &= 5x &\text p - 5xa &= 5c \\ p &= 5(xa + c) \end{align}$$

This implies that both $p$ and $q$ are divisible by $5$. Which contradicts GCD($p$,$q$) = $1$ since $5$ is a greater common divisor than $1$. Thus the assumption of $a + \sqrt[b]5$ is rational is false.

$\endgroup$
9
  • 1
    $\begingroup$ first, assume that the gcd of p and q is 1, which you can always do...but make sure to note that explicitly. Second, are you talking about $b\cdot \sqrt 5$ or $b^{\sqrt 5}?$ $\endgroup$
    – Alan
    Commented Oct 12, 2014 at 1:08
  • $\begingroup$ I edited the confusing part sorry about that. Now check if it makes sense. Thanks $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 5:46
  • 1
    $\begingroup$ where you have $5q^b=5(5^{b-1}c^b)$, that does not imply q is a multiple of 5...since you can cancel a 5 from each side, what you have left with is $q^b=5^{b-1}c^b$. This would only mean q is a multiple of 5 IF $b-1\ge 1$, but unfortunately $b-1$ can be 0 $\endgroup$
    – Alan
    Commented Oct 12, 2014 at 6:30
  • $\begingroup$ Thanks Alan I will look into it and try to fix it $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 8:01
  • 1
    $\begingroup$ To make this easier, $a + \sqrt[b]5$ is irrational if and only if $\sqrt[b]5$ is irrational. $\endgroup$
    – DanielV
    Commented Oct 12, 2014 at 8:30

2 Answers 2

3
$\begingroup$

Your last paragraph is wrong:

This implies that both p and q are multiples of 5 so there is a contradiction because the equation is irrational so there cannot be any common factors for it to be irrational.

It's like you've seen a proof of contradiction before, borrowed some words from it, and had no idea what the actual logic was. Memorizing won't get you far in mathematics. It looks like random incoherent rambling.

To correct your proof, first, your statement

For some integer p,q where q is not zero

is incomplete. You need to specify that $\frac pq$ is in lowest common terms, in other words that $\operatorname{GCD}(p,q) = 1$.

Second, your conclusion should be:

  • Therefore $p$ and $q$ are both divisible by $5$
  • This contradicts the assumption that $\operatorname{GCD}(p,q) = 1$
  • Therefore the assumption that $a + \sqrt[b]5$ is rational must be false
$\endgroup$
6
  • $\begingroup$ I actually wrote this proof myself I just didn't understand how to conclude it. Also what does GCD?. Is it greatest common denominator? $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 7:51
  • $\begingroup$ I appreciate the help but the way you described my mistake of the conclusion isn't the most polite way. $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 7:53
  • $\begingroup$ See en.wikipedia.org/wiki/Greatest_common_divisor and en.wikipedia.org/wiki/Irreducible_fraction . I wasn't trying to be mean, but it really does not seem like a language/grammar issue. I suggest organizing your thoughts into multiple sentences instead of trying to make one long sentence. $\endgroup$
    – DanielV
    Commented Oct 12, 2014 at 7:57
  • $\begingroup$ Alan stated my line "5^qb=5(5^(b−1)c^b)" is not true when x=0 but in your response you did not mention it being wrong. So does that mean it works? $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 7:58
  • 1
    $\begingroup$ Alan is correct. Consider whether $a + \sqrt[1]5$ irrational. I missed that you wrote $b \ne 0$ rather than $b > 1$ and he caught it. $\endgroup$
    – DanielV
    Commented Oct 12, 2014 at 8:02
1
$\begingroup$
  1. $b \in \mathbb{Z^+}\\ \sqrt[b]5=\dfrac{p}{q}, \operatorname{gcd}(p,q)=1\\\implies 5=\dfrac{p^b}{q^b} \\ \implies 5q^b=p^b\\ \implies p \mid 5 \\\implies p=5 \\\implies q^b=5^{b-1}$

$b-1>0 \implies 5\mid q \implies q=5k \implies 5^b \cdot k^b=5^{b-1} \implies 5k^b=1$

$\therefore \not \exists p,q\mid \operatorname{gcd}(p,q)=1 \land \sqrt[b]5=\dfrac{p}{q}$ ,$\forall b \in \mathbb{N}$ $\implies$ $\sqrt[b]5 \in \mathbb {R} \setminus \mathbb{Q} \implies a+\sqrt[b]5 \in \mathbb {R} \setminus \mathbb{Q}$ , $\forall a \in \mathbb {Z}$

$b \in \mathbb{Z^-} \implies \exists t\in \mathbb {N} \mid b=-t \implies \sqrt[b]5=\sqrt[-t]5=\dfrac{1}{5^{\dfrac{1}{t}}}$

$\exists r,s \mid \operatorname{gcd}(r,s)=1, t \in\mathbb {N} \land \sqrt[-t]5=\dfrac{r}{s} \implies \sqrt[t]5 = \dfrac{s}{r}$ !!

$\endgroup$
8
  • $\begingroup$ So then what I did is right I just have to prove for b-1 > 0 correct? But b-1 can also be equal to 0 why did you not include that? $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 18:41
  • $\begingroup$ @geforce: I have included that. Please read the argument once more. $\endgroup$
    – user170039
    Commented Oct 12, 2014 at 18:42
  • $\begingroup$ OK I see, u used if b is a member of negative integers and solved it that way $\endgroup$
    – geforce
    Commented Oct 12, 2014 at 18:55
  • $\begingroup$ Double check $b=1 \implies \sqrt{5} \in \mathbb{R} \setminus \mathbb{Q}$. It is vacuously true, but what you should be concerned of is $\sqrt[1]{5} \in \mathbb{R} \setminus \mathbb{Q}$ $\endgroup$
    – DanielV
    Commented Oct 12, 2014 at 19:45
  • $\begingroup$ @DanielV: Thanks for your suggestion. Removed it. $\endgroup$
    – user170039
    Commented Oct 13, 2014 at 3:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .