Is a + (1/√2)(b-a) Irrational?

[charmedbeauty]

Homework Statement

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

a+(1/√2)(b-a) is irrational.

Homework Equations

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.

 

[Mentallic]

so a+(b-a)/√2 =p/q

a(1+(b-1)/√2) = p/q

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

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

so (p/qa) -1= (b-1)/√2

qa/p-1= √2/(b-1)

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.

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.

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.

 

[charmedbeauty]

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.

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.

 

[SammyS]

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.

That looks much better !

 

[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.

 

[charmedbeauty]

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.

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.

 

[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.

 

[charmedbeauty]

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 ([itex](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, [itex]2n^2= (2k)^2= 4k^2[itex]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.[/itex][/itex][/itex]

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

 

[BruceW]

...for integers , m and n where m and n have no common factors.

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).

So in a test situation would it be necessary to prove that √2 is irrational as a sought of "side proof"?

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.

 

[charmedbeauty]

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.

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

 

[HallsofIvy]

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).

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

 

[BruceW]

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

 

[charmedbeauty]

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

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

 

[Bohrok]

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

 

[SteveL27]

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

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