Proof that non-integer root of an integer is irrational

In summary, the conversation was about a possible proof for the statement that the square root of an integer is either an integer or an irrational number. The conversation included a proposed proof which was later found to be flawed, and a second proposed proof which the other person needed some time to understand. The summary also includes clarification on the scope of the statement being proved and an explanation of the second proposed proof.
  • #1
Swamp Thing
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I have been looking at various proofs of this statement, for example Proof 1 on this page : http://www.cut-the-knot.org/proofs/sq_root.shtml

I'd like to know if the following can be considered as a valid and rigorous proof:

Given ##y \in \mathbb{Z}##, we are looking for integers m and n ##\in \mathbb{Z}## such that ##(m/n)^2=y##.

We can write ##m/n=x## where ##x \notin \mathbb{Z}##.
Thus ##(m/n)^2=x^2=y##, or
##m^2=x^2n^2##

##m^2-x^2n^2=0##

##(m-xn)(m+xn)=0##

Taking only the positive value,
##m=xn##
Now ##x \notin \mathbb{Z}##,
which means ##xn \notin \mathbb{Z}## which implies ##m \notin \mathbb{Z}##
which contradicts our requirement that ##m\in \mathbb{Z}##

So no pair of m, n can be found as specified.

===

Edit : We could just go directly to ##m=xn## from ##(m/n)^2=x^2=y## , perhaps without loss of rigor?
 
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  • #2
The assertion that nx is not an integer because x is not an integer is false.
i.e. x=1.5 (not an integer) and n=2 (an integer) nx=3 (an integer).
 
  • #3
Silly me! :cry:

Thank you.
 
  • #4
No worries - the best of us makes these sorts of mistakes. It is why peer-review and 3rd party oversight is such a big part of research.
 
  • #5
Actually, I had (what I thought was) a proof, but more complicated than the one I posted. In trying to simplifiy it, I got to a point where it became "simpler than possible".

After re-checking version one, I'll post it here for your comments.
 
  • #6
Swamp Thing said:
Given y∈Zy \in \mathbb{Z}, we are looking for integers m and n ∈Z\in \mathbb{Z} such that (m/n)2=y(m/n)^2=y.
So if I give you y = 9, is it impossible to find m and n?
 
  • #7
Svein said:
So if I give you y = 9, is it impossible to find m and n?

The statement to be proved specifically excludes this kind of example from its scope, since it refers to non-integer roots.

Actually a better way to write it would be, "the square root of an integer is either an integer or an irrational number".

But that wouldn't have fitted in the subject box so I shortened it a bit.
 
  • #8
I can give you the start of a proof, but I haven't finished it (that's where the professor leaves the rest for the student to do).

  1. Given n∈ℤ, assume [itex]n=(\frac{p}{q})^{2} [/itex] where p, q ∈ℤ.
  2. In addition assume that p and q have no common factors (otherwise you could divide both p and q with that factor without changing the value of the fraction).
  3. In addition assume q>1 (to make it a proper fraction)
Now, if [itex]n=(\frac{p}{q})^{2} [/itex] then [itex]\frac{p^{2}}{n\cdot q^{2}}=1 [/itex]. Since p and q have no common factors, n must divide p2. Then either n is a square by itself (n = m2) and therefore divides p, or it is not a square.

In the first case p = r⋅m (r, m ∈ℤ), but since p and q have no common factors, you cannot have [itex]\frac{p^{2}}{n\cdot q^{2}}=\frac{r^{2}\cdot m^{2}}{m^{2}\cdot q^{2}}=\frac{r^{2}}{q^{2}} = 1[/itex].

Now it is up to you to handle the case where n is not a square.
 
  • #9
Thanks! I'll need a while to get my head around that.

But wait.. the statement to be proved is about n's that are NOT squares (see post #7) so you have left me with all the work still to do! :frown:
 
  • #10
Swamp Thing said:
Thanks! I'll need a while to get my head around that.

But wait.. the statement to be proved is about n's that are NOT squares (see post #7) so you have left me with all the work still to do! :frown:
It is not that hard! Since p and q have no common factors, p2 and q2 have no common factors. Then look at [itex] \frac{p^{2}}{n\cdot q^{2}}[/itex]. Here either n divides p2 or it does not. In the first case ([itex]p^{2}=n\cdot u [/itex]) you end up with [itex]\frac{p^{2}}{n\cdot q^{2}}=\frac{n\cdot u}{n \cdot q^{2}}=\frac{u}{q^{2}} [/itex] and again, since p and q have no common factors, this expression cannot be 1.

Second case: n does not divide p2. Then nominator and denominator in [itex]\frac{p^{2}}{n\cdot q^{2}} [/itex] have no common factors and therefore the expression cannot be 1.
 
  • #11
Thanks, I get it now!
 

Related to Proof that non-integer root of an integer is irrational

What is the definition of an integer?

An integer is a whole number that can be positive, negative, or zero. It does not contain any fractions or decimals.

What is an irrational number?

An irrational number is a number that cannot be expressed as a fraction of two integers. It is a non-terminating and non-repeating decimal.

Why is the proof of a non-integer root of an integer being irrational important?

This proof is important because it helps us understand the properties of irrational numbers and their relationship to integers. It also allows us to prove that certain numbers, such as the square root of 2, are irrational.

How is the proof of a non-integer root of an integer being irrational demonstrated?

The proof involves assuming that the root is a rational number and then reaching a contradiction. This can be done by showing that the assumption leads to a contradiction with the fundamental properties of integers and rational numbers.

What are some real-world applications of this proof?

This proof has various applications in fields such as engineering, physics, and computer science. For example, it is used in signal processing to design digital filters and in cryptography for generating secure codes.

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