Square Root of an Irrational Number is Irrational

[hammonjj]

Homework Statement

Let a be a positive real number. Prove that if a is irrational, then √a is
irrational. Is the converse true?

Homework Equations

So, an irrational number is one in which m=q/p does not exist. I understand that part, but then trying to show that the square root of an irrational number is irrational is giving me problems. I would think this needs to be done by contradicition

Thanks!

 

[LCKurtz]

Homework Statement

Let a be a positive real number. Prove that if a is irrational, then √a is
irrational. Is the converse true?

Homework Equations

So, an irrational number is one in which m=q/p does not exist.

More carefully, what you mean is that there are not integers p and q such that m = q/p.

I understand that part, but then trying to show that the square root of an irrational number is irrational is giving me problems. I would think this needs to be done by contradicition

Thanks!

I would think so to. So try it...

 

[hammonjj]

Here's what I have so far:

Proceed by way of contradiction. Assume a is rational, so a=x/y. So

√a=p/q
√(x/y)=p/q
x/y=p^2/q^2
q^2(x/y)=p^2

This is where I don't know where to go from here. Normally, I would take advaadvantagetnage of knowing something about a, that is was prime or even for example, but I don't know what I an say to show that there is indeed a contradiction.

Thanks

 

[Dick]

Here's what I have so far:

Proceed by way of contradiction. Assume a is rational, so a=x/y. So

√a=p/q
√(x/y)=p/q
x/y=p^2/q^2
q^2(x/y)=p^2

This is where I don't know where to go from here. Normally, I would take advaadvantagetnage of knowing something about a, that is was prime or even for example, but I don't know what I an say to show that there is indeed a contradiction.

Thanks

Which part are you trying prove?

 

[hammonjj]

Which part are you trying prove?

I'm trying to show that if a is a rational number then √a can not be rational (the contradiction). Like I said in the OP, I'm pretty lost with this one.

 

[hammonjj]

Could I say a=√2, which is an irrational number?

 

[Dick]

I'm trying to show that if a is a rational number then √a can not be rational (the contradiction). Like I said in the OP, I'm pretty lost with this one.

That's not really the contradiction. And you can't prove it because it's not true. Can't you think of a simple counterexample? What's the correct contradiction (or contrapositive)?

 

[Dick]

I'm trying to show that if a is a rational number then √a can not be rational (the contradiction). Like I said in the OP, I'm pretty lost with this one.

4 is rational, so is √4=2. The contradiction of "if a is irrational, then √a is
irrational" is "if √a is NOT irrational then a is NOT irrational".

 

[alanlu]

4 is rational, so is √4=2. The contradiction of "if a is irrational, then √a is
irrational" is "if √a is NOT irrational then a is NOT irrational".

You mean contrapositive, right? A proof by contradiction would be like "if √a is not irrational and a is rational, then (deductive logic), and thus (clearly false statement), but (point out why the statement is false) and thus contradiction."

But yeah, use the contrapositive.

 

[Dick]

You mean contrapositive, right? A proof by contradiction would be like "if √a is not irrational and a is rational, then (deductive logic), and thus (clearly false statement), but (point out why the statement is false) and thus contradiction."

But yeah, use the contrapositive.

Yes, of course, I meant contrapositive. I said that in post 7. I was trying to follow the OP's words and may have muddled this up more than I helped. Thanks.

 

[rktpro]

It could be like this. First you prove that something like √2 is irrational.
Then let, on contrary, √(√2) = p/q where p and q are co-prime
Square both sides, √2= p^2/q^2=(p/q)^2
But you know that the square of any fraction which contains co-prime can't be irrational or something with an under root.
So, this contradiction has arise because our assumption was wrong.

 

[SammyS]

By contradiction.

Suppose that a is irrational but that √(a) is rational.

Then √(a) that \displaystyle \sqrt{a}=\frac{p}{q}\,, for some integers p and q .

Now consider what all this says about (\sqrt{a})^2\,.

 

[swades]

let √a be rational
√a=p/q
this √a,being a rational no. must be between 2 integers,so
I< p/q <I+1
=Iq<p<q(I+1)
=0<p-Iq<q

now,as p and q are integers , therefore (p-Iq) is also an integer,so
when it is multiplied by a rational no. ,it will be a rational no i.e.
k= (p-Iq)*(p/q)...(integer)*(rational no), this will be a rational no.
now,k=(p^2)/q -Ip
also,k={(p^2)/(q^2)}q - Ip
then,k=aq - Ip........as (p^2)/(q^2)=a
now,k=(irrational)*integer - integer
hence k is also irrational,thereby arriving at a contradiction,where before it was
supposed to be rational .. ..
so ,therefore √a is also irrational when a is rational

 

[HallsofIvy]

This

this √a,being a rational no. must be between 2 integers, so I< p/q <I+1

Is NOT true. If √a= 3 it does not lie between I and I+1 for any integer I.

Why not simply "if \sqrt{a}= m/n[/tex]&quot; then a= what?

 

[swades]

let √a be rational,(including integers)
√a=p/q
this √a,being a rational no. must be between any 2 consecutive integers,so
I≤ p/q <I+1
=Iq≤p<q(I+1)
=0≤p-Iq<q
now let us take, p- Iq=0
so,I=p/q=√a
also,I^2=a ,which makes "a"(which is irrational) ,an integer.hence a contradiction.
so,excluding the equality case,we have
=0<p-Iq<q......(A)
now,as p and q are integers , therefore (p-Iq) is also an integer(lets say I1),so
when it is multiplied by a rational no. ,it will be a rational no i.e.
k= (p-Iq)*(p/q)...{integer(I1)}*(a rational no,whose denominator is an integer less than I1)......from (A)
this "k" will be a rational no.(since q cannot be a factor of I1)
now,k=(p^2)/q -Ip
also,k={(p^2)/(q^2)}q - Ip
then,k=aq - Ip........as (p^2)/(q^2)=a
now,k=(irrational)*integer - integer
hence k is also irrational,thereby arriving at a contradiction,where before it was
supposed to be rational .. ..
so ,therefore √a is also irrational when a is rational

hope ,this time its correct...

 

[Dick]

let √a be rational,(including integers)
√a=p/q
this √a,being a rational no. must be between any 2 consecutive integers,so
I≤ p/q <I+1
=Iq≤p<q(I+1)
=0≤p-Iq<q
now let us take, p- Iq=0
so,I=p/q=√a
also,I^2=a ,which makes "a"(which is irrational) ,an integer.hence a contradiction.
so,excluding the equality case,we have
=0<p-Iq<q......(A)
now,as p and q are integers , therefore (p-Iq) is also an integer(lets say I1),so
when it is multiplied by a rational no. ,it will be a rational no i.e.
k= (p-Iq)*(p/q)...{integer(I1)}*(a rational no,whose denominator is an integer less than I1)......from (A)
this "k" will be a rational no.(since q cannot be a factor of I1)
now,k=(p^2)/q -Ip
also,k={(p^2)/(q^2)}q - Ip
then,k=aq - Ip........as (p^2)/(q^2)=a
now,k=(irrational)*integer - integer
hence k is also irrational,thereby arriving at a contradiction,where before it was
supposed to be rational .. ..
so ,therefore √a is also irrational when a is rational

hope ,this time its correct...

Why are you making this so complicated? If sqrt(a)=p/q, then sure a=p^2/q^2. Which is rational. So if a is irrational, then mustn't it be that sqrt(a) is irrational? It's just a simple proof by contradiction. I don't know what all the extra complication is about.

 

[D H]

Why are you making this so complicated? If sqrt(a)=p/q, then sure a=p^2/q^2. Which is rational. So if a is irrational, then mustn't it be that sqrt(a) is irrational? It's just a simple proof by contradiction.

That's proof by contraposition: Prove ⌉q→⌉p and you have proved p→q. Proof by contradiction would involve showing that the statement "there exists an irrational number a whose square root is rational" results in a contradiction.

 

[Dick]

That's proof by contraposition: Prove ⌉q→⌉p and you have proved p→q. Proof by contradiction would involve showing that the statement "there exists an irrational number a whose square root is rational" results in a contradiction.

I think it's a fine distinction. "there exists an irrational number a whose square root is rational" implies a is rational. Isn't that a contradiction?

 

[D H]

Sure. But you don't need to contradiction to prove this. All one has to prove is that if √a is rational then a is a rational. By contraposition, if a is irrational then √a is irrational. QED. There's no need to look at the inverse of "if a is irrational then √a is irrational" and find a contradiction because the two statements "if √a is rational then a is rational" and "if a is irrational then √a is irrational" are one and the same, just worded differently.

 

[1MileCrash]

The way I would do this is prove that the square root of any non-perfect square is irrational. This is easily done with prime factorization in a couple of lines.

Then, since the definition of a perfect square requires that it is a natural number, the proof is complete.

 

[1MileCrash]

I see that there seems to be an ongoing discussion here regarding.. something about the second part of the question?

"Prove that if a is irrational, sqrt(a) is irrational. Is the converse true?"

The converse of that statement is:
If sqrt(a) is irrational, a is irrational.

Counter example: any rational number a that is not a perfect square (say, 2)

So no, the converse is not true, proof complete.

 

[Dick]

The way I would do this is prove that the square root of any non-perfect square is irrational. This is easily done with prime factorization in a couple of lines.

Then, since the definition of a perfect square requires that it is a natural number, the proof is complete.

I don't see what that is supposed to prove, but it sure doesn't prove the square root of an irrational is irrational.

 

[1MileCrash]

I don't see what that is supposed to prove, but it sure doesn't prove the square root of an irrational is irrational.

Right, I think that method would only prove it for non-perfect square naturals, my mistake.

So, it's even easier

If sqrt(a) is rational, a is rational

Is the contrapositive, which is trivially true because you can't square a rational and get anything but another rational. So there is no proof by contradiction, it's just a direct proof.

 

[Dick]

Right, I think that method would only prove it for non-perfect square naturals, my mistake.

So, it's even easier

If sqrt(a) is rational, a is rational

Is the contrapositive, which is trivially true because you can't square a rational and get anything but another rational. So there is no proof by contradiction, it's just a direct proof.

You can prove it either by proving the contrapositive or proving the original statement by contradiction, they are both the same thing really. It's a matter of taste, I think.

 

[1MileCrash]

I do agree that proof by contradiction and contrapositive are related in a formulaic way, but in this case, since the contrapositive is immediately apparent due to a closure property, I would forgo a proof by contradiction purely for simplicity and the length of the proof.

Then again, that's only if you're in the habit of writing the contrapositive down along with the theorem (I personally do because sometimes it helps my intuition.)

 

[swades]

Why are you making this so complicated? If sqrt(a)=p/q, then sure a=p^2/q^2. Which is rational. So if a is irrational, then mustn't it be that sqrt(a) is irrational? It's just a simple proof by contradiction. I don't know what all the extra complication is about.

on logical and common reasoning it might be simple to say that root of an irrational no. will be irrational.
But what mathematics requires is a well defined way of proving a statement.
BOLZANO WEIERSTRAUS theorem states that "every bounded & infinite set has a limit point".
On a superficial basis it is easy to say this ,but proving the same will require elegant elaboration.
even basic mathematics have got their own twists and turns.

I know that mine was not an elegant one.yet i want to know whether it has even 1% correctness &genuineness in it or not.

 

[D H]

To be honest, your approach is very long, very convoluted, and to make matters worse it is poorly written. Finding the problem, if one exists, is not an exercise I want to undertake. This is a simple exercise. A few lines should do it.