Proving the Irrationality of Square Root of 3

[r0bHadz]

Homework Statement

Prove sqrt(3) is irrational

Homework Equations

The Attempt at a Solution

(a/b)^2 = 3 assume a/b is in lowest form

a^2 = 3b^2

so a^2 is of form 3n

whenever n is even, a^2 will be even => a will be even whenever n is even

so a is of form 2l whenever n is even

=> 4l^2 = 3b^2 => 2l/sqrt(3) = b

rewrite as 2((l*sqrt(3))/3) = b

contradiction. Since both a and b are divisible by 2 whenever n is even, sqrt(3) must be irrational for even n, This implies that sqrt(3) is irrational

 

[fresh_42]

Homework Statement

Prove sqrt(3) is irrational

Homework Equations

The Attempt at a Solution

(a/b)^2 = 3 assume a/b is in lowest form

a^2 = 3b^2

so a^2 is of form 3n

whenever n is even, a^2 will be even => a will be even whenever n is even

so a is of form 2l whenever n is even

=> 4l^2 = 3b^2 => 2l/sqrt(3) = b

rewrite as 2((l*sqrt(3))/3) = b

contradiction.

To what?

Since both a and b are divisible by 2 whenever n is even, sqrt(3) must be irrational for even n, This implies that sqrt(3) is irrational

What if $n$ is odd?

I think you can do it a lot easier. Do you know what a prime number is, and that $3$ is prime?

 

[r0bHadz]

To what?

What if $n$ is odd?

I think you can do it a lot easier. Do you know what a prime number is, and that $3$ is prime?

Contradicts a/b being a rational in lowest form.

I don't see why proving its irrational for even n to not be sufficient enough...

But yes I know that 3 is prime

 

[fresh_42]

Contradicts a/b being a rational in lowest form.

I don't see why proving its irrational for even n to not be sufficient enough...

But yes I know that 3 is prime

Why should it be sufficient? $n$ odd is a possibility, or not? And I do not see any contradiction, because all you know (by assumption) that $\sqrt{3}$ is rational, so what follows from an expression like $\dfrac{2\cdot l \cdot \sqrt{3}}{3}=b$ other than $b \in \mathbb{Q}$.

What about my other question: Do you know what a prime number is?

 

[r0bHadz]

Why should it be sufficient? $n$ odd is a possibility, or not? And I do not see any contradiction, because all you know (by assumption) that $\sqrt{3}$ is rational, so what follows from an expression like $\dfrac{2\cdot l \cdot \sqrt{3}}{3}=b$ other than $b \in \mathbb{Q}$.

What about my other question: Do you know what a prime number is?

Hmm I guess you're right. I'll continue with the possibility of n being odd..

As for the contradiction. a^2 = 3b^2 => a^2 is of form 3n => a is of form 3n. If a is of form 3n => 3b^n that b will be even as well, and since I assumed (a/b) is in lowest form there is a contradiction since I have two evens numbers.

Yes a prime number is one whose factors are only 1 and itself. I think you are saying it will be easier to solve this problem if I can show that (a/b)^2 = 3 that if b does not equal one then 3 is not a prime so I would have a contradiction?

 

[fresh_42]

As for the contradiction. a^2 = 3b^2 => a^2 is of form 3n => a is of form 3n. If a is of form 3n => 3b^n that b will be even as well, and since I assumed (a/b) is in lowest form there is a contradiction since I have two evens numbers.

I don't see this.

Yes a prime number is one whose factors are only 1 and itself. I think you are saying it will be easier to solve this problem if I can show that (a/b)^2 = 3 that if b does not equal one then 3 is not a prime so I would have a contradiction?

That is wrong. Your definition is the one for an irreducible number, not a prime number. For the integers the two terms are identical, i.e. the prime numbers are exactly the irreducible ones and vice versa. However, the true definition of a prime number is of a great help here:

A number is $p$ is prime, if it is not $\pm 1$ and if $p \,|\,a\cdot b \Longrightarrow p\,|\,a \text{ or } p\,|\,b$. With this definition, the proof is more or less straightforward and you don't need cases and expressions at the end, which are composed by elements you want to examine: $\sqrt{3}\,.$

Another possibility is to use the prime factor decompositions of $a$ and $b$.

 

[r0bHadz]

I don't see this.

That is wrong. Your definition is the one for an irreducible number, not a prime number. For the integers the two terms are identical, i.e. the prime numbers are exactly the irreducible ones and vice versa. However, the true definition of a prime number is of a great help here:

A number is $p$ is prime, if it is not $\pm 1$ and if $p \,|\,a\cdot b \Longrightarrow p\,|\,a \text{ or } p\,|\,b$. With this definition, the proof is more or less straightforward and you don't need cases and expressions at the end, which are composed by elements you want to examine: $\sqrt{3}\,.$

Another possibility is to use the prime factor decompositions of $a$ and $b$.

Sorry I do not know why I wrote that in the first quote, when that is not what I meant. I meant to write:

a^2 = 3b^2 => if b^2 is even, then a^2 is also even, => a and b are both even, thus (a/b) cannot be in lowest form.

I will try to finish this problem using the definition of prime.

 

[fresh_42]

a^2 = 3b^2 => if b^2 is even, then a^2 is also even, => a and b are both even, thus (a/b) cannot be in lowest form.

This is correct. But the case $b$ odd could be a problem.

I will try to finish this problem using the definition of prime.

Just start with $3\,|\,a^2$ and so $3\,|\,a$ and so on.

 

[member 587159]

I don't see this.

That is wrong. Your definition is the one for an irreducible number, not a prime number. For the integers the two terms are identical, i.e. the prime numbers are exactly the irreducible ones and vice versa. However, the true definition of a prime number is of a great help here:

A number is $p$ is prime, if it is not $\pm 1$ and if $p \,|\,a\cdot b \Longrightarrow p\,|\,a \text{ or } p\,|\,b$. With this definition, the proof is more or less straightforward and you don't need cases and expressions at the end, which are composed by elements you want to examine: $\sqrt{3}\,.$

Another possibility is to use the prime factor decompositions of $a$ and $b$.

You can't call a definition wrong when it is equivalent with the one you have in mind, even if yours is more general as it works in the more general settings of rings. The context was very clear here.

 

[fresh_42]

You can't call a definition wrong when it is equivalent with the one you have in mind, even if yours is more general as it works in the more general settings of rings. The context was very clear here.

I can call it wrong because it is. That they are equivalent is a theorem, not given a priori. I don't like it, if students are taught mistakes. Me, too, has learned it wrong at school, but what is the matter with doing it right? It has nothing to do with "more general", it is a different property.

You won't define compact by bounded and closed, just because it is the same in $\mathbb{R}^n$, won't you?

 

[member 587159]

I can call it wrong because it is. That they are equivalent is a theorem, not given a priori. I don't like it, if students are taught mistakes. Me, too, has learned it wrong at school, but what is the matter with doing it right? It has nothing to do with "more general", it is a different property.

You won't define compact by bounded and closed, just because it is the same in $\mathbb{R}^n$, won't you?

It is perfectly fine to define compactness in $\mathbb{R}^n$ as closed and bounded. In fact, there are a couple of books that do this. The author can then prove it is equivalent with the cover definition and then use this as general definition in topological/metric spaces. Whether this is pedagogically sound, is another question. But definitely not wrong. The reason this is done is that covers are hard to grasp at first, so giving an alternative "more intuitive" definition helps students.

In the same way, I think the definition of the OP is more intuitive than the one you propose. Probably the reason that most books use this definition. A quick search also shows that wikipedia uses the OP's definition.

 

[fresh_42]

Sorry, but I'm against the method of doing something wrong and correct it afterwards. It implies the hidden assumption that students / kids are not smart enough to do it correctly. Make a test and start a thread by: "1. Given $X\subseteq \mathbb{R}^n$ compact, show that ... 2. Compact = bounded and closed 3. ..." And use a statement which is easy to proof be covering compactness and hard by the other one, as it is here the case.
I bet it won't take two posts, until this definition will be corrected here!

Sorry, but irreducibility is not prime! That it is the same for some rings, doesn't justify to introduce it wrong and correct it afterwards. I hate this didactic method: "(1st year) Today we learn addition and subtraction. ... No, 3-5 is not possible, you cannot give away 5 bonbons if you have only 3." and then "(some time later) Today we learn that 3-5 = -2" I really do hate this method!

Irreducibility is not prime! Stop assuming kids are stupid.

 

[fresh_42]

And even if I accepted the wrong definition at schools, that is no reason to accept it on PF!

 

[member 587159]

Sorry, but I'm against the method of doing something wrong and correct it afterwards. It implies the hidden assumption that students / kids are not smart enough to do it correctly. Make a test and start a thread by: "1. Given $X\subseteq \mathbb{R}^n$ compact, show that ... 2. Compact = bounded and closed 3. ..." And use a statement which is easy to proof be covering compactness and hard by the other one, as it is here the case.
I bet it won't take two posts, until this definition will be corrected here!

Sorry, but irreducibility is not prime! That it is the same for some rings, doesn't justify to introduce it wrong and correct it afterwards. I hate this didactic method: "(1st year) Today we learn addition and subtraction. ... No, 3-5 is not possible, you cannot give away 5 bonbons if you have only 3." and then "(some time later) Today we learn that 3-5 = -2" I really do hate this method!

Irreducibility is not prime! Stop assuming kids are stupid.

It isn't a matter of thinking kids are stupid.

Why not start with general topological spaces and then treat metric spaces as a special case? Simply because people wouldn't see the point.

Or maybe, let us start with measure theory before doing probability theory.

Sometimes, you must make people confident in a certain area, way of thinking before they can appreciate/understand the whole picture. This has nothing to do with being stupid or not.

That's why we teach kids things IN SMALL STEPS. One thing at a time.

 

[fresh_42]

I do not agree. You contradict yourself:

Sometimes, you must make people confident in a certain area, way of thinking before they can appreciate/understand the whole picture.

It isn't a matter of thinking kids are stupid.

The former statement implies that you know what's building up confidence and what does not. This is a hidden way of saying too complicated which in return hides too stupid.

But apart from this discussion about pedagogic: Do you really think irreducibility is necessary to understand primality? I may agree with you on measure theory, although I'm less convinced than you are, but I do not agree, that primality is significantly more difficult than irreducibility. The comparison with measure theory is a bit far fetched.

 

[member 587159]

I do not agree. You contradict yourself:The former statement implies that you know what's building up confidence and what does not. This is a hidden way of saying too complicated which in return hides too stupid.

But apart from this discussion about pedagogic: Do you really think irreducibility is necessary to understand primality? I may agree with you on measure theory, although I'm less convinced than you are, but I do not agree, that primality is significantly more difficult than irreducibility. The comparison with measure theory is a bit far fetched.

I think $p|ab \implies p|a \lor p|b$ is more difficult to comprehend than the alternate definition, yes. But that wasn't the point. The point was that both definitions are good (at least in this context!). And maybe you are right that there are mathematical reasons to prefer the one over the other, but to call it wrong is simply wrong itself. Mathematically, the OP gave a correct definition (again, in this context).

I disagree that I contradicted myself. "Too complicated" can mean "too soon (YET)" or "not mature enough (YET)". Not "too stupid to understand".

Another example: starting with calculus before diving in real analysis. It illustrates the case well imo.

Let us agree to disagree.

 

[fresh_42]

Let us agree to disagree.

Agreed. The more as I've chosen "wrong" as adjective to break the automatism with irreducibility rather than to start a debate upon its linguistic implications.

 

[r0bHadz]

a^2 = 3b^2 => a = 3((b^2)/a) => 3|a

so a is of form 3n

9n^2 = 3((b^2)
=>
3n^2 = b^2 => 3((n^2)/b) = b => 3|b

So (a/b) isn't in lowest form, therefore sqrt3 is irrational

 

[r0bHadz]

Fresh would this proof be sufficient enough?

 

[fresh_42]

a^2 = 3b^2 => a = 3((b^2)/a) => 3|a

so a is of form 3n

9n^2 = 3((b^2)
=>
3n^2 = b^2 => 3((n^2)/b) = b => 3|b

I would avoid fractions here, as it requires to demonstrate that $b\,|\,n^2$ which needs a second thought. Why is $\dfrac{n^2}{b}$ an integer?

Instead you could just repeat the primality argument: $9n^2=3b^2 \Longrightarrow 3n^2=b^2 \Longrightarrow 3\,|\,b$ and if three divides both, you are done.

So (a/b) isn't in lowest form, therefore sqrt3 is irrational

 

[r0bHadz]

I would avoid fractions here, as it requires to demonstrate that $b\,|\,n^2$ which needs a second thought. Why is $\dfrac{n^2}{b}$ an integer?

Instead you could just repeat the primality argument: $9n^2=3b^2 \Longrightarrow 3n^2=b^2 \Longrightarrow 3\,|\,b$ and if three divides both, you are done.

Hmm okay

I would avoid fractions here, as it requires to demonstrate that $b\,|\,n^2$ which needs a second thought. Why is $\dfrac{n^2}{b}$ an integer?

Instead you could just repeat the primality argument: $9n^2=3b^2 \Longrightarrow 3n^2=b^2 \Longrightarrow 3\,|\,b$ and if three divides both, you are done.

Ah good point. I never would have caught that. Alright I appreciate it. I am going to do the same for sqrt 5 now, ill report back if I have any difficulties

 

[fresh_42]

One last question. We know if 3|a^2 => 3|a from the following theorem

if x|f => x|fo for all o ∈ ℤ and its converse is true

Not really. It follows from $3$ being prime. A prime number which divides a product has to divide a factor of the product. It is the correct definition of prime: $3\,|\,a\cdot a \Longrightarrow 3\,|\,a$.

E.g. $3\,|\,20\cdot 24$ then either $3\,|\,20$ or $3\,|\,24$. If we have a non prime, say $30$ then $30\,|\,20\cdot 24 =480$ but $30\nmid 20$ and $30\nmid 24$ because the prime factors of $30$ are distributed among $20$ and $24$.

For the proof you want to write, it is the convenient property. If you would have defined prime as irreducible number, i.e. there are no proper factors in $3$, then it was not obvious from the definition for this irreducibility, how $3\,|\,a^2$ implies $3\,|\,a$. Why should this be?

As mentioned earlier, you can use the prime factor decomposition as well. Assume $a=p_1^{r^1}\cdot \ldots \cdot p_m^{r_m},$ then $3\,|\,a^2=p_1^{2r^1}\cdot \ldots \cdot p_m^{2r_m}$ implies that one of the factors $p_i = 3\,.$ and thus $3\,|\,a$.