# Prove this is irrational.

1. Apr 26, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

Prove that if a and b are rational numbers with a≠b then

a+(1/√2)(b-a) is irrational.

2. Relevant equations

3. The attempt at a solution

Assume that a+(1/√2)(b-a) is rational.

then by definition of rationality

a+(1/√2)(b-a) =p/q for some integers p&q

so a+(b-a)/√2 =p/q

a(1+(b-1)/√2) = p/q

so (p/qa) -1= (b-1)/√2

qa/p-1= √2/(b-1)

so √2 = (b-1)((qa/p) -1)

but (b-1)((qa/p) -1) is rational since b,1,q,a,p are all integers and the sum, difference and products of integers are integers.

but √2 is not an integer. Contradiction a+(1/√2)(b-a) must be irrational.

2. Apr 26, 2012

### Mentallic

This step is incorrect.

If $$a+\frac{b-a}{\sqrt{2}}=\frac{p}{q}$$

then in order to factorize out a from the LHS, you'll need a factor of a in each and every term, but the LHS can also equivalently be written like this:

$$a+\frac{b}{\sqrt{2}}-\frac{a}{\sqrt{2}}$$

So factoring out a would look like this:

$$a\left(1+\frac{b}{a\sqrt{2}}-\frac{1}{\sqrt{2}}\right)$$

and not

$$a\left(1+\frac{b-1}{\sqrt{2}}\right)=a\left(1+\frac{b}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)$$

This step is also incorrect. If we have something of the form

$$\frac{a}{b}+c$$
then the reciprocal of that is not $$\frac{b}{a}+c$$

Take that extra minute to do the steps in between to find out what it is.

The logic to answer the question is correct though, you just need to fix up your algebra.

Oh and all you need to do is solve for $\sqrt{2}$ so factorng out a is a pointless step.

3. Apr 26, 2012

### charmedbeauty

Thanks a bunch Mentallic!

so keeping the logic the same the algebra should look a little more like this...

a+ (1/√2)(b-a)=p/q

(1/√2)(b-a)=(p/q)-a

√2 = (b-a)/((p/q)-a)

but (b-a)/((p/q)-a) is rational....Hence a+ (1/√2)(b-a) is irrational.

end of proof.

4. Apr 26, 2012

### SammyS

Staff Emeritus
That looks much better !

5. Apr 26, 2012

### HallsofIvy

What properties of rational numbers are you allowed to use?
The rational numbers are closed under addition and subtraction so the first thing I would have done is say "if $a+ (b- a)/\sqrt{2}= r$, a rational number, then $(b- a)/\sqrt{2}= r- a$ is rational. Since $b\ne a$, $b- a\ne 0$ and, since the rational numbers are closed under division by any number other than 0, $1/\sqrt{2}= (r- a)/(b- a)$ is rational.

Finally, since 1 over any number is not 0 and b- a is not 0, $b- a/\sqrt{2}$ is not 0 so r- a is not 0, so we would have to conclude that $\sqrt{2}= (b- a)/(r- a)$ is rational. Now, if you already know that $\sqrt{2}$ is not rational, that sufficient. If you do not already have that, it is fairly easy to prove just as you were progressing.

6. Apr 27, 2012

### charmedbeauty

That's the thing that confuses me with these proofs... Is it fair to say that √2 is irrational without proving it, because it is sought of a tautology.

7. Apr 27, 2012

### HallsofIvy

I said "Now, if you already know that $\sqrt{2}$ is not rational, that sufficient. If you do not already have that, it is fairly easy to prove just as you were progressing."
It is "fair to say that $\sqrt{2}$ is irrational" if that has already been proven. If not, as I said, it is easier to prove than this number, using the same ideas as you did: if $\sqrt{2}$ were rational, we could write $\sqrt{2}= m/n$ for integers , m and n where m and n have no common factors. Squaring, $2= m^2/n^2$ and so $m^2= 2n^2$. The square of any odd number is odd ($(2i+1)^2= 4i^2+ 4i+ 1= 2(2i^2+ 2i)+ 1)$ so since $m^2$ is even, m must be even. That is, m= 2k for some integer k. In that case, $2n^2= (2k)^2= 4k^2$ and so $n^2= 2k^2$. Since $n^2$ is even, n must also be even, contradicting the fact that m and n have no common factors.

Last edited by a moderator: Apr 28, 2012
8. Apr 27, 2012

### charmedbeauty

Ok thanks but the original question does not tell you if √2 is rational or irrational. It just asks about the number involving the √2... So in a test situation would it be necessary to prove that √2 is irrational as a sought of "side proof"?

9. Apr 27, 2012

### BruceW

But then you would also need to prove that any ratio of two integers can be re-written as a ratio of two integers with no common factors. (since a rational number is defined as being a ratio of any two integers, with the denominator being non-zero).

Good question. I would think it's sufficient to just say 'and the square root of two is irrational'. But I only did a physics degree, so I am not familiar with how mathematical proofs are examined.

10. Apr 27, 2012

### charmedbeauty

Thanks BruceW... I have noticed that they overly anal about some things thoe!

11. Apr 28, 2012

### HallsofIvy

Every text I have seen defines a rational number as the ratio of two integer, having no common factors.

12. Apr 28, 2012

### BruceW

really? so 6/3 is not a rational number? I'm not a mathematician, so I don't know these things...

13. Apr 28, 2012

### charmedbeauty

it seems no one has a definition for irrational numbers... not even mathematicians:surprised

14. Apr 28, 2012

### Bohrok

I think he means the ratio of two integers is in its most reduced form have no common factor...

15. Apr 29, 2012

### SteveL27

I'm not sure I believe that. In fact I'm sure I don't believe that. Because the "no common factors" condition is superflous to the definition. As someone else pointed out, 6/3 is a perfectly good rational number. I don't need to reduce it to lowest terms to see that it's rational, because it's the ratio of two integers. Period.

Not that "proof by Wiki" is the last word, but they agree with this point.

a rational number is any number that can be expressed as the quotient or fraction a/b of two integers, with the denominator b not equal to zero.

http://en.wikipedia.org/wiki/Rational_number