~~Proving that people still discover things about the oldests of proofs, this is from Geoffrey C. Berresford in 2008.~~ [Claims abound of it being much older, so bonus kudos to whoever finds the oldest reference.]

It is almost stupidly easy (and hence is very smart).

**Prove**: If is a rational number then is a perfect square.

**Suppose**: is a rational number in lowest terms.

Doing a little algebra…

… we can say

Since is in lowest terms, there’s an integer such that and .

Since we also know . That is, is an integer, so is an integer and is a perfect square.

Taking the contrapositive, if is a not a perfect square then is not a rational number.

Filed under: Mathematics |

cheesemonkeysf, on May 25, 2011 at 11:10 am said:Very nice!

Jason Dyer, on May 25, 2011 at 12:12 pm said:A colleague (Jan Wehr) pointed out to me this is related to a more compact proof…

Let B / A be lowest rational form of sqrt(N).

B / A = sqrt(N) / 1

B^2 / A^2 = N / 1

Since B and A share no factors B^2 / A^2 is also in lowest terms. N/1 is clearly in lowest terms. So B^2 = N, that is, N is a perfect square.

erichawk, on July 29, 2021 at 5:29 pm said:I think there is a problem with this proof. Squaring B / A = sqrt(N) / 1 and concluding that B^2 / A^2 = N / 1 and then equating the denominator or one fraction as the multiple of another (A^2 * c = 1, so that both A and c = 1) asuumes N is an integer. It should instead be written as B^2 / A^2 = N_B / N_A, where N_B and N_A are integers. Continuing as before yields B^2 * c = N_B and A^2 * c = N_A, which doesn’t help.

Consider similar reasoning, replacing sqrt(N) with x:

x / 1 = 3 / 2 (lowest form expression of the fraction)

x^2 / 1 = 9 / 4 (this is still in lowest form, as you note)

But it’s not valid to conclude from this that there is some integer c such that

9 * c = x^2

and

4 * c = 1

(There is no such integer c. Indeed, if we attempt to conclude that c = 1 from this and use that result, we would also conclude that

9 * 1 = x^2

and

x = 3,

which is clearly not true, as we stated x = 3 / 2 at the outset).

erichawk, on July 29, 2021 at 5:41 pm said:Or it is assuming that sqrt(n) is an integer, so that it can be rewritten as sqrt(n) / 1.

Alexander Bogomolny, on May 25, 2011 at 2:41 pm said:This is a proof by Geoffrey C. Berresford (Am Math Monthly, Vol. 115, No. 6 (June-July 2008), p. 524). For more, see

http://www.cut-the-knot.org/proofs/sq_root.shtml

Jason Dyer, on May 25, 2011 at 9:04 pm said:That link (and the author credit) is in the post already. But it’s ok to repeat it here I guess.

LSquared, on May 25, 2011 at 4:54 pm said:when I was a TA at UC Riverside about a decade ago the professor who taught pre-calc always did this for sqrt(2) in one of his early lectures.

gasstationwithoutpumps, on May 25, 2011 at 9:16 pm said:I learned that proof in the late 1960s, and it was considered an old technique even then. Why someone gets author credit for failing to do library research and reinventing the wheel is beyond me.

Jason Dyer, on May 25, 2011 at 9:20 pm said:Knowing how things in math get lost and found again I don’t doubt this at all, but I’d love a source so I can fix up the post.

gasstationwithoutpumps, on May 25, 2011 at 10:41 pm said:I think the proof goes back to the Greeks. Here is a citation from the 1800s:

http://books.google.com/books?id=hgoAAAAAYAAJ&pg=PA130&dq=proof+that+square+root+of+two+is+irrational&hl=en&ei=F-bdTcWmIcrRiALenr3VCg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CEkQ6AEwBQ#v=onepage&q&f=false

Ray’s algebra: designed for high schools and colleges. An analytical …, Part 2

Art. 179 page 130

(1852)

Jason Dyer, on May 26, 2011 at 5:59 am said:That’s like my proof in the comments, not the first (supposedly recent) proof.

It’s only subtly different for you and me, but the latter [as Jan pointed out when he mentioned the proof] feels more to beginners like something mystical.

gasstationwithoutpumps, on May 26, 2011 at 8:48 am said:The subtle differences seem to be differences in writing style, which has changed a bit over the past 160 years, rather than differences in the math, so I’d argue that it is indeed the same proof, and pretty much the same proof that the Greeks used (though I don’t have any classic Greek to read the original).

Jason Dyer, on May 26, 2011 at 8:57 am said:The second proof relies on “when squaring the numerator and denominator which previously did not have common factors still do not have common factors and is therefore still irreducible”. The first proof instead relies on “if the fractions are in the same equivalence class there is a multiple that makes the numerators and denominators equal”. This is nontrivial.

Did you check out the rest of that book you linked to? It’s actually pretty neat.

Ben Blum-Smith, on May 29, 2011 at 9:05 pm said:I second Jason. These are not the same proof. The proof quoted in the book relies on the fundamental theorem of arithmetic, but the proof given by Jason only relies on the claim “X/Y in lowest terms and X/Y = Z/W implies there is integer c with Z=cX and W=cY.” They probably generalize a bit differently. If not, the equivalence doesn’t seem trivial to me.

MathNewbie, on May 26, 2011 at 5:26 pm said:“Since B/A is in lowest terms, there’s an integer c such that Bc=Na and ac=b.”

Sorry I don’t understand your proof (I’m a student learning about proofs), but why is there an integer c? Aren’t you assuming what you’re trying to prove?

Thanks.

Jason Dyer, on May 26, 2011 at 8:02 pm said:Just the raw statement

B/A = NA/B

implies that the fractions are equivalent, but not that the numerators and denominators are equal, because the fractions might be, say 1/2 and 2/4. However, we do know when fractions are equivalent and one of them is in lowest terms we can multiply the lowest-terms fraction by some integer to get the other fraction. So 1/2 = 2/4 but we can do 1 * 2 / 2 * 2 to get 2/4 = 2/4. Or 2/5 = 6/15 but we can multiple 2 * 3 / 5 * 3 to get 6/15. This only works when one of the fractions is in lowest terms, though; think of 2/4 and 3/6.

So where did c come from? Well, since we know the fraction property is true for some integer, we can peg that integer at c. We haven’t assumed anything about c other than it is “some integer” so we aren’t assuming anything beyond what we can.

Now if you’re wanting a proof of that property, it’s fair to ask for one, but it’s considered to be one of those “already proved” things in most proofs of the irrationality of square roots, just like we can assume the fundamental theorem of arithmetic (which I proved a few posts ago) is also true.

MathNewbie, on May 27, 2011 at 11:55 pm said:Aha! Thanks for the reply. I understand it now.

studentOfLife, on October 17, 2012 at 8:52 am said:This is genius. I love the mathematical thinking behind this proof. I am starting to see the beauty of proofs, where at first I just hated them because I did not know where to start chipping away the problem.

Thanks

Anonymous, on April 5, 2013 at 3:12 pm said:I can provide a counter example sqrt(59)