- #1

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- 593

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?

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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