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Proving sqrt of 3 is irrational

  • Thread starter armolinasf
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Homework Statement



So, I'm trying to prove that the square root of 3 is irrational.

2. Attempt at a Solution

2x can be any even number and 2x+1 can be any odd number. Since an irrational number is any number that can't be expressed as a ratio of two integers, I just have to show that the ratio of any two integers squared, either odd/even, even/odd, or odd/odd will not equal 3.

I think I've proved that any odd/even or even/odd cannot be the square root of three but I'm having trouble with odd/odd.

For odd/even and even/odd:

(2x+1)/(2x)=sqrt(3) ==> (2x+1)=sqrt(3)(2x)

square both sides:

4x^2+4x+1=12x^2 ==>even+1 does not equal an even.

However if I apply the same method with two odd numbers say (2x+1)/(2y+1)=sqrt(3). get the following:

4x^2+4x+1=12x^2+12x+3

I'm not sure where to go from there...Thanks in advance for the help
 

Answers and Replies

  • #2
Mentallic
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Your logic is flawed. For the first case where you have odd/even, using (2x+1)/2x implies that for any integer x, the denominator is 2x and the numerator is one more than that. For example, 3/2, 5/4, 13/12 etc. What about all other cases?
By the way you don't need to consider cases where the numerator and denominator are odd or even, but rather just two separate integers.

Also, what about (2x+1)/(2x)=sqrt(9), squaring both sides give 4x^2+4x+1=36x^2 so with the same logic you used, since the left side is odd and the right side is even sqrt(9) is irrational. This obviously isn't true.

Start with a/b=sqrt(3) and thus a^2=3b^2. Now if b is odd then b^2 is odd and if b is even then b^2 is even (you can try prove this too if you like). See if you can continue from there.
 
  • #3
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(2x+1)/(2y+1)=sqrt(3). this is right, but don't substitute x for y

you can manipulate this to get 4x^2+4x+1=12y^2+12y+3, move some terms around and factor out a 2 to get 2x^2 + 2x - 6y^2 - 6y = 1, the left side will always be even, thus you've established your inequality

as for mentallic's idea, I think I've seen that proof for sqrt of 2 somewhere but it's a bit fuzzy, feel free to expand on that method
 
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  • #4
Your proof with work as long as you keep the numerator and denominator of your fraction arbitrary.

You made an error in using [itex] \frac{2x+1}{2x}[/itex] as mentallic pointed out.

You should have used [itex] \frac{2x+1}{2y}[/itex].

mentallic suggestion would also shorten your proof.
 
  • #5
Delta2
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You should have used y instead of x for the denominator only(or the numerator only) but other than that proof remains same. hy23 is correct for the rest of proof.

Something else relative to this. All these proofs are based on the fact that the set of natural numbers is partitioned to two subsets, the set of even and the set of odds. Dont strart throwing rocks on me (lol) but how do you prove that (i.e that every natural number is of the form 2n or 2m+1).. Hmm induction principle?
 
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  • #6
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Okay so 4x^2+4x+1=12y^2 and 4y^2=12x^2+12x+3 would not work? Since as Mentallic said:

Also, what about (2x+1)/(2x)=sqrt(9), squaring both sides give 4x^2+4x+1=36x^2 so with the same logic you used, since the left side is odd and the right side is even sqrt(9) is irrational. This obviously isn't true.

So if I take x/y=sqrt3 then x^2=3y^2.

if x is even and y is odd we have an odd number times and odd number and that won't be even. And if x is odd and y is even an even times and odd will not produce an even number.

But wouldn't this face the same argument raised by the sqrt9 example?

But if both x and y are odd then following hy's example 2x^2+2x-6y^2+6y=1
 
  • #7
HallsofIvy
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I don't think "even" and "odd" is the right way to go about this, since you are talking about "3" rather than "2". I would be inclined to go with "a multiple of 3", "a multiple of 3 plus 1", or "a multiple of 3 plus 2".

In any case, if [itex]\sqrt{3}= m/n[/itex], then [itex]m^2 3n^2[/itex] so that m^2is a multiple of 3. You should be able to show, by looking at [itex](3a+ 1)^2[/itex] and [itex](3a+ 2)^2[/itex], that m itself must be a multiple of 3. And, of course, if [itex]m= 3p[/itex] then [itex]m^2= 9p^2= 3n^2[/itex] so that [itex]n^2= 3p^2[/itex].
 
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  • #8
Delta2
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In the example with sqrt(9) all one proves by taking (2x+1)/(2x) is that sqrt(9) cannot be the ratio (2x+1)/(2x) for some number x. Ok thats good but it is not exactly what we want and this doesnt prove that sqrt(9) is irrational.
 
  • #9
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Can you explain a little bit more why a multiple of three might work better? I'm not quite following...
 
  • #10
Mentallic
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In the example with sqrt(9) all one proves by taking (2x+1)/(2x) is that sqrt(9) cannot be the ratio (2x+1)/(2x) for some number x. Ok thats good but it is not exactly what we want and this doesnt prove that sqrt(9) is irrational.
Well of course it doesn't prove root(9) is irrational, because it isn't. I think you meant root(3) here. And like I said earlier, using (2x+1)/2x isn't correct as x has to be some integer, thus the numerator is always 1 more than the denominator. Use (2x+1)/2y instead.
 
  • #11
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but using (2x+1)/2y would be correct? How does that differ with using multiples of 3?
 
  • #12
Delta2
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but using (2x+1)/2y would be correct? How does that differ with using multiples of 3?
Yes would be correct.

Using multiples of 3 you just go to prove that if x/y=sqrt(3) then x and y are both multiples of 3 that is x=3k, y=3p which contradicts the initial hypothesis which you can make from start that the fraction x/y is irreducible (which means that x and y dont have commmon divisor other than 1).
 
  • #13
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Useful hint: All the prime factors of a square number above 1, can always be sorted in pairs where every pair contains two equal prime factors. Only square numbers have this ability.
For instance: 12² = (2·2·3)² = 2·2 times 2·2 times 3·3
That means if a square number is divisible by a prime number p, it is also divisible by p².
(For instance, if a square number is divsible by 5, it is also divisible by 5²=25.)

I would suggest using √3=x/y instead of √3=(2x+1)/(2y). Maybe you'll see how it's better to use multiples of 3 by solving it this way, and by taking a look at the hint above.
 
  • #14
Mentallic
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Yes would be correct.

Using multiples of 3 you just go to prove that if x/y=sqrt(3) then x and y are both multiples of 3 that is x=3k, y=3p which contradicts the initial hypothesis which you can make from start that the fraction x/y is irreducible (which means that x and y dont have commmon divisor other than 1).
I'm sorry but how can you make the assumption that x and y are both multiples of 3?
I don't see what you've tried to prove.
 
  • #15
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I don't think "even" and "odd" is the right way to go about this, since you are talking about "3" rather than "2". I would be inclined to go with "a multiple of 3", "a multiple of 3 plus 1", or "a multiple of 3 plus 2".

In any case, if [itex]\sqrt{3}= m/n[/itex], then [itex]m^2 3n^2[/itex] so that m^2is a multiple of 3. You should be able to show, by looking at [itex](3a+ 1)^2[/itex] and [itex](3a+ 2)^2[/itex], that m itself must be a multiple of 3. And, of course, if [itex]m= 3p[/itex] then [itex]m^2= 9p^2= 3n^2[/itex] so that [itex]n^2= 3p^2[/itex].
That's what I'm wondering too. I saw this approach in Gelfand's algebra but I never really understood why you would use multiples of 3...

Im curious if we can't prove the sqrt of 9 isn't irrational by contradiction then how can we prove the sqrt of 2 or 3 by contradiction?
 
  • #16
Delta2
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I'm sorry but how can you make the assumption that x and y are both multiples of 3?
I don't see what you've tried to prove.
Er i mean i make the initial assumption that x and y arent multiples of 3 (actually the assumption that they dont have any other common devisor except 1. This assumption just means that the fraction x/y is in simplified form with all the common divisors taken out, there is nothing wrong with it, every fraction can be written in simplified form e.g 90/100 is 9/10 or 27/30 is again 9/10).

Then by x/y=sqrt(3) you can prove that both x and y are multiples of 3. [tex]x^2=3y^2[/tex] which means [tex]x^2[/tex] is a multiple of 3. If [tex]x^2[/tex] is multiple of 3 then you can prove that x is also multiple of 3 (if x=3k+1 or x=3k+2 then it turns out that [tex]x^2[/tex] isnt multiple of 3). Since x=3k then [tex]x^2=9k^2=3y^2[/tex] hence [tex]y^2=3k^2[/tex] hence y^2 and hence y is also multiple of 3. So we proved that x and y are both multiples of 3 which contradicts our initial assumption.
 
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  • #17
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So what you're saying is that if we use a multiple of three we have another way to prove by contradiction that sqrt3 is irrational.

But that means we can still use 2x+1 and 2y as a legitimate way of proving it as well right?
 
  • #18
HallsofIvy
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Buy since it is the square root of 3 you are dealing with, why be concerned with 2x+ 1 or 2y at all?

[itex](3n)^2= 9n^2= 3(3n^2)[/itex] so a multiple of 3, squared, is again a multiple of 3.

[itex](3n+ 1)^2= 9n^2+ 6n+ 1= 3(3n^2+ 2n)+ 1[/itex] so any number that is 1 more than a multiple of 3 is the same, squared.

[itex](3n+2)^2= 9n^2+ 12n+ 4= 3(3n^3+ 4n+ 1)+ 1[/itex] so any number that is 2 more than a multiple of 3 is 1 more than 3, squared.
 
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  • #19
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I'm not sure if I'm following. What does this tell us about how the numbers relate to the square root of three?

In the last to instances, (3n+1)^2 and (3n+2)^2, I can see how if they are both one greater than a multiple of three squared, does that mean since they are one greater that the sqrt of three cannot be expressed in terms of such a number?
 

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