Proof that non-integer root of an integer is irrational

  • #1
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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Answers and Replies

  • #2
Simon Bridge
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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).
 
  • #4
Simon Bridge
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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
Svein
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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
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
Svein
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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
Svein
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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!
 

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