Proving √3 is irrational

[Gromit57]

I'm working through the book Book of Proof (3rd edition) by Richard Hammack (freely available, BTW).
Chapter 6, exercise 5 (page 144) asks to prove that √3 is irrational.
The proof I came up with is different than the book's solution, and other solutions I found online, so I'm not sure it's correct.

Suppose for contradiction that √3 is rational.
Then √3 = a/b, for some integers a, b.
Assume a/b is fully reduced.
Squaring both sides and multiplying by b² gives a² = 3b².

a² is either odd or even.
Suppose a² is even, then a is even (this was shown in a previous exercise).
Then b is odd (since if b was even, both a and b would have a common factor of 2, and we assumed a/b is fully reduced).
Then there's an integer c such that a = 2c, and plugging that into the first equation gives 4c² = 3b².
Then 3b² is even, which means b is even (since if b was odd, then b² would be odd, and 3b² would be a product of two odds, which itself is odd).
But we deduced that b is odd, so we get a contradiction.

Suppose a² is odd, then a is odd, so b is even.
Then b = 2c for some integer c, and plugging that into the first equation gives a² = 3(2c)² = 12c².
So a² is even, but we assumed a² is odd, so we get a contradiction.

So a² is neither odd nor even, which is a contradiction.
Therefore, √3 is irrational.

Is this correct?

 

[Deleted member 4993]

I'm working through the book Book of Proof (3rd edition) by Richard Hammack (freely available, BTW).
Chapter 6, exercise 5 (page 144) asks to prove that √3 is irrational.
The proof I came up with is different than the book's solution, and other solutions I found online, so I'm not sure it's correct.

Suppose for contradiction that √3 is rational.
Then √3 = a/b, for some integers a, b.
Assume a/b is fully reduced.
Squaring both sides and multiplying by b² gives a² = 3b².

a² is either odd or even.
Suppose a² is even, then a is even (this was shown in a previous exercise).
Then b is odd (since if b was even, both a and b would have a common factor of 2, and we assumed a/b is fully reduced).
Then there's an integer c such that a = 2c, and plugging that into the first equation gives 4c² = 3b².
Then 3b² is even, which means b is even (since if b was odd, then b² would be odd, and 3b² would be a product of two odds, which itself is odd).
But we deduced that b is odd, so we get a contradiction.

Suppose a² is odd, then a is odd, so b is even.
Then b = 2c for some integer c, and plugging that into the first equation gives a² = 3(2c)² = 12c².
So a² is even, but we assumed a² is odd, so we get a contradiction.

So a² is neither odd nor even, which is a contradiction.
Therefore, √3 is irrational.

Is this correct?

You said:

Suppose a² is odd, then a is odd, so b is even. - not correct! a and b could be prime-numbers!

An approximation of \(\displaystyle \sqrt{3}\) could be \(\displaystyle \frac{433}{251}\)

Here both the numbers are odd (like 5 and 7) and those are relatively prime.

Good exercise though...

 

[lev888]

Suppose a² is odd, then a is odd, so b is even.

Why?

 

[Gromit57]

Why?

You're right, they could both be odd.

Thank you.

 

[Steven G]

You're right, they could both be odd.

It is true that if a^2 is odd then a is odd. Now we know a^2 = 3b^2. Since a^2 is odd then 3b^2 is odd. 3 is clearly odd. Now an odd*odd is odd and an odd*even is even. So b^2 must be odd. But then b is odd.
Assuming a is odd, to say that both a and b could both be odd is not exactly correct. Rather (assuming a is odd) a and b both MUST be odd.

 

[Gromit57]

Rather (assuming a is odd) a and b both MUST be odd.

Thank you for pointing this out.

Unfortunately, using the fact that b is odd, so that b = 2c + 1 for some integer c, and plugging that into the first equation gives a² = 3(2c + 1)² = ... = 2(6c² + 6c + 1) + 1, which means a is odd, which doesn't contradict anything.

 

[firemath]

What an "odd" problem!

 

[Steven G]

Suppose \(\displaystyle \sqrt{3}= \frac {a}{b}\) in reduced form.
Then \(\displaystyle a^2 = 3b^2\)
Now \(\displaystyle a^2\) is a multiple of 3\(\displaystyle \implies \)a is a multiple of 3 (PROVE THIS!)
So \(\displaystyle a = 3k \) and then \(\displaystyle a^2=9k^2\)
Now we have \(\displaystyle 9k^2 = 3b^2 \implies\) b is a multiple of 3.
There is your contradiction.

In proving \(\displaystyle \sqrt{2} \) is irrational you need to get a and b to be both multiples of 2 (ie show that a and b are even)
In proving \(\displaystyle \sqrt{3} \) is irrational you need to get a and b to be both multiples of 3
In proving \(\displaystyle \sqrt{5} \) is irrational you need to get a and b to be both multiples of 5

You really need to prove that if \(\displaystyle a^2\) is a multiple of 3, then so is \(\displaystyle a\).
If \(\displaystyle a^2\) is a multiple of 5, then so is \(\displaystyle a\).

Please post back with your work that I requested.

 

[Steven G]

What an "odd" problem!

Do you know what the oddest prime is?

 

[lev888]

Do you know what the oddest prime is?

?

 

[Gromit57]

Suppose \(\displaystyle \sqrt{3}= \frac {a}{b}\) in reduced form.
Then \(\displaystyle a^2 = 3b^2\)
Now \(\displaystyle a^2\) is a multiple of 3\(\displaystyle \implies \)a is a multiple of 3 (PROVE THIS!)
So \(\displaystyle a = 3k \) and then \(\displaystyle a^2=9k^2\)
Now we have \(\displaystyle 9k^2 = 3b^2 \implies\) b is a multiple of 3.
There is your contradiction.

Yes, this is the book's solution.

 

[Steven G]

Yes, this is the book's solution.

Yes, this is the standard method. I knew that your proof had to be wrong because you were talking about even and odd and not multiples of 3

 

[pka]

I'm working through the book Book of Proof (3rd edition) by Richard Hammack (freely available, BTW).
Chapter 6, exercise 5 (page 144) asks to prove that √3 is irrational.
The proof I came up with is different than the book's solution, and other solutions I found online, so I'm not sure it's correct

Because you are using Hammack's work you surely can follow this proof.
Suppose that \(\displaystyle P\in\mathbb{Z}^+\) and \(\displaystyle (\forall n\in\mathbb{Z}[P\ne n^2)\)(i.e. P is not a square).
You should know about the floor function: thus \(\displaystyle 0 < \sqrt P - \left\lfloor {\sqrt P } \right\rfloor < 1\) (**)
O.K. here goes: Suppose that \(\displaystyle \sqrt P\) is a rational number.
Then there are two relatively prime positive integers \(\displaystyle a~\&~b\) such that \(\displaystyle \sqrt{P}=\dfrac{a}{b}\) or \(\displaystyle a=b\sqrt{P}\)
Define a set \(\displaystyle T=\{n\in\mathbb{Z}^+: n\sqrt{P}\in\mathbb{Z}^+\}\). \(\displaystyle T\) is not empty because \(\displaystyle b\in T\)
So \(\displaystyle T\) is a non-empty subset of the positive integers as such \(\displaystyle T\) contains a first term. Call it \(\displaystyle k\).
From (**) we get \(\displaystyle 0<k\sqrt P-kH<k\). BUT \(\displaystyle (k\sqrt P-kH)\in\mathbb{Z}^+\). Why??
Moreover, \(\displaystyle (k\sqrt P-kH)<k\) and \(\displaystyle (k\sqrt P-kH)(\sqrt P)\in\mathbb{Z}^+\).
That violates the minimality of \(\displaystyle k\),. Its a contradiction to supposing \(\displaystyle \sqrt P\) is rational.

To restate: If \(\displaystyle P\) is a positive integer that is not a square then \(\displaystyle \sqrt P\) is irrational.

 

[Dr.Peterson]

Thank you for pointing this out.

Unfortunately, using the fact that b is odd, so that b = 2c + 1 for some integer c, and plugging that into the first equation gives a² = 3(2c + 1)² = ... = 2(6c² + 6c + 1) + 1, which means a is odd, which doesn't contradict anything.

I presume that at this point last night, your question had been answered, as you now see why your attempted proof is not successful. You already know the correct proof, as you stated from the start:

Chapter 6, exercise 5 (page 144) asks to prove that √3 is irrational.
The proof I came up with is different than the book's solution, and other solutions I found online, so I'm not sure it's correct.
...
Is this correct?