# Fixing a Calculation Mistake on p.51 in “Mathematical Analysis I” by Vladimir A. Zorich?

While reading "Mathematical Analysis I" by Vladimir A. Zorich. I came across this proof regarding the irrationality of $$\sqrt{2}$$ :

Let $$X$$ and $$Y$$ be the sets of positive real numbers such that $$\forall x \in X\left(x^{2}<2\right)$$ $$\forall y \in Y\left(2 Since $$1 \in X$$ and $$2 \in Y,$$ it follows that $$X$$ and $$Y$$ are nonempty sets. Further, since $$(x for positive numbers $$x$$ and $$y,$$ every element of $$X$$ is less than every element of $$Y .$$ By the completeness axiom there exists $$s \in \mathbb{R}$$ such that $$x \leq s \leq y$$ for all $$x \in X$$ and all $$y \in Y$$ We shall show that $$s^{2}=2$$ If $$s^{2}<2,$$ then, for example, the number $$s+\frac{2-s^{2}}{3 s},$$ which is larger than $$s,$$ would have a square less than $$2 .$$ Indeed, we know that $$1 \in X,$$ so that $$1^{2} \leq s^{2}<2,$$ and $$0<\Delta:=2-s^{2} \leq 1 .$$ It follows that :

$$\boxed{\left(s+\frac{\Delta}{3 s}\right)^{2}=s^{2}+2 \cdot \frac{\Delta}{3 s}+\left(\frac{\Delta}{3 s}\right)^{2}

The author made a simple mistake during the expansion as seen in the box above. Therefore, I attempt to fix the proof as follow :

Since $$1 \leq s^2$$ and $$\frac{\Delta}{3}<1$$ hold, then : \begin{align*} \left(s+\frac{\Delta}{3s}\right)^2&=s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3s}\right)^2\\ &\leq s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3}\right)^2\\ &
Hence, $$(s+\frac{\Delta}{3s})^2<2$$ is true as required.

Are my steps correct in my attempt to fix this part of the proof?

• Well seems you've corrected it, it is just a typo that the 3 in the expression should be 2. – macton Jan 24 at 5:54
• I agree looks correct to me too. – xX A C E Xx Jan 24 at 5:57
• @macton Thank you very much. – tchappy ha Jan 24 at 6:18
• @xXACEXx Thank you very much. – tchappy ha Jan 24 at 6:19
• @xXACEXx Thank you very very much for your edit. – tchappy ha Jan 24 at 9:02

What year was your edition of Zorich published? A copy of the Russian edition from 2012 is here. It has the expansion you write about on the bottom of page number 77 and the calculations there match your correction.

• KCd, I am reading Vladimir A. Zorich "Mathematical Analysis I Second Edition". And this book was translated by Roger Cooke and Octavio Paniagua T. – tchappy ha Jan 24 at 9:07
• KCd, thank you very much for your answer. – tchappy ha Jan 24 at 9:08
• I bought a copy of the book from Springer in late July last year. – tchappy ha Jan 24 at 9:11
• I found "Preface to the Sixth Russian Edition" in this book. So, I guess this book was translated from Sixth Russian Edition. – tchappy ha Jan 24 at 9:19
• That link to a Russian version of the book goes to the 6th edition: the first page at the link says Издание шестое and you can throw that into Google Translate to see it means "sixth edition". So what you found is a printing error in the translation. – KCd Jan 24 at 9:23

Alternate approach:

Worth looking at the concrete equation $$x^2 - 2 = 0$$ we are dealing with. We know that there are methods to solve this equation. Among them, there is the Newton approximation (going on the tangent), and the secant method. In this case, since we want values less than the solution, we use the secant method.

So consider the points on the graph $$(s, s^2-2)$$ below the horizontal, axis, and $$(2, 2^2 - 2)$$ above the horizontal axis. Join them and consider the point of intersection with the horizontal axis $$(s', 0)$$. From the picture, we should have $$s'> s$$, and $$s'^2-2 < 0$$.

Now, a simple calculation gives $$s' = \frac{2(s+1)}{s+2}$$ Check that $$\begin{eqnarray} s' - s &=& \frac{2-s^2}{s+2}\\ 2 - s'^2 &=& \frac{2(2-s^2)}{(s+2)^2} \end{eqnarray}$$

• orangeskid, Thank you very much for your answer. – tchappy ha Jan 24 at 9:04
• @tchappy ha: you are very welcome! Good luck in your study! – orangeskid Jan 24 at 9:31