Proving the Irrationality of Square Root of 3

  • Thread starter r0bHadz
  • Start date
  • Tags
    Irrational
In summary, to prove that ##\sqrt{3}## is irrational, one can use the definition of a prime number and follow the proof by contradiction method. By assuming that ##\sqrt{3}## is rational, one can express it as a ratio of two integers, ##a/b##, in lowest form. By squaring both sides, it is shown that ##3## is a factor of ##a^2##, which implies that ##3## is also a factor of ##a##. This leads to a contradiction, as both ##a## and ##b## would have a common factor of ##3##, contradicting the assumption that ##a/b## is in lowest form. Therefore, ##\sqrt{
  • #1
r0bHadz
194
17

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
 
Physics news on Phys.org
  • #2
r0bHadz said:

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?
 
  • #3
fresh_42 said:
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
 
  • #4
r0bHadz said:
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?
 
  • #5
fresh_42 said:
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?
 
  • #6
r0bHadz said:
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##.
 
  • #7
fresh_42 said:
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.
 
  • #8
r0bHadz said:
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.
 
  • #9
fresh_42 said:
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.
 
  • #10
Math_QED said:
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?
 
  • #11
fresh_42 said:
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.
 
  • Like
Likes PeroK
  • #12
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.
 
  • #13
And even if I accepted the wrong definition at schools, that is no reason to accept it on PF!
 
  • #14
fresh_42 said:
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.
 
  • Like
Likes PeroK
  • #15
I do not agree. You contradict yourself:
Math_QED said:
Sometimes, you must make people confident in a certain area, way of thinking before they can appreciate/understand the whole picture.
Math_QED said:
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.
 
  • #16
fresh_42 said:
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.
 
  • #17
Math_QED said:
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.
 
  • #18
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
 
  • Like
Likes fresh_42
  • #19
Fresh would this proof be sufficient enough?
 
  • #20
r0bHadz said:
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
 
  • #21
fresh_42 said:
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
fresh_42 said:
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
 
  • #22
r0bHadz said:
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##.
 

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers. In other words, it cannot be written as a fraction where the numerator and denominator are both whole numbers. Irrational numbers include numbers like pi and the square root of 2.

2. How do you prove that sqrt(3) is irrational?

There are a few different ways to prove that sqrt(3) is irrational. One way is to assume that sqrt(3) is rational and then use proof by contradiction to show that this assumption leads to a contradiction. Another way is to use the rational root theorem, which states that if a polynomial has rational roots, those roots must be integers. Since the polynomial x^2 - 3 has no integer roots, sqrt(3) must be irrational.

3. Can you give an example of a proof by contradiction for sqrt(3) being irrational?

Yes, here is one example of a proof by contradiction for sqrt(3) being irrational:
Assume that sqrt(3) is rational and can be expressed as a fraction a/b, where a and b are integers with no common factors.
Then, we can square both sides to get 3 = a^2/b^2.
Multiplying both sides by b^2, we get 3b^2 = a^2.
This means that a^2 is divisible by 3, which means that a must also be divisible by 3.
But if a is divisible by 3, then a^2 is divisible by 9.
This means that 3b^2 is divisible by 9, which means that b^2 must also be divisible by 3.
But if b^2 is divisible by 3, then b must also be divisible by 3.
This contradicts our initial assumption that a and b have no common factors.
Therefore, our assumption that sqrt(3) is rational must be false, and thus sqrt(3) is irrational.

4. Can you prove that any number with a non-perfect square root is irrational?

Yes, the proof for any number with a non-perfect square root being irrational follows a similar logic to the proof for sqrt(3) being irrational. We can assume that the number can be expressed as a fraction a/b, where a and b are integers with no common factors. Then, we can square both sides and use proof by contradiction to show that this assumption leads to a contradiction. Therefore, any number with a non-perfect square root is irrational.

5. Why is proving that sqrt(3) is irrational important in mathematics?

Proving that sqrt(3) is irrational is important because it helps us understand the properties of numbers and their relationships with each other. It also helps us develop more complex mathematical concepts and theories. Additionally, this proof can be applied to other numbers with non-perfect square roots, leading to a deeper understanding of irrational numbers and their significance in mathematics.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
1
Views
983
  • Precalculus Mathematics Homework Help
Replies
4
Views
770
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
735
  • Precalculus Mathematics Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Precalculus Mathematics Homework Help
Replies
21
Views
705
  • Precalculus Mathematics Homework Help
Replies
13
Views
2K
Back
Top