# Why do these seemingly identical calculations give different answers?

Consider the two equations:

$$\dfrac{3}{2}\sqrt 2\approx \dfrac{543}{256}$$ and $$\dfrac{3}{\sqrt 2} \approx \dfrac{543}{256}$$

The left-hand sides of both are equivalent so it seems like they should yield the same value when approximating $$\sqrt 2$$.

For the first equation: $$\sqrt 2 \approx \dfrac{543}{256} \cdot \dfrac{2}{3} = \dfrac{181}{128}$$

For the second equation: $$\sqrt 2 \approx 3 \cdot \dfrac{256}{543}= \dfrac{256}{181}$$

The results are close in value but not the same.

Why do these methods give different results?

• Well, it might be true that $\sqrt{2} \approx x$ and $\sqrt{2} \approx y$ while $x \not = y$. Mar 5, 2019 at 21:22
• But in this case each equation uses the same starting value of $\frac{543}{256}$ so why should $x \neq y$? Mar 5, 2019 at 21:25
• If I assume that $5 = 5.001$ then equivalently $0 = 0.001$ and equivalently $0 = 1$. Doing arbitrary transformations doesn't guarantee to keep approximation good enough. Mar 5, 2019 at 21:29
• Because taking reciprocals need not preserve $\approx$ to the same accuracy. Your approximations are only "identical" when one overlooks that they are applied to $\sqrt 2$ and its reciprocal. Mar 5, 2019 at 21:33
• "But in this case each equation uses the same starting value of 543/256 so why should x≠y?" But in both cases that starting value is wrong. So we should manipulations remain consistent? Mar 5, 2019 at 22:28

Think of this problem from a perspective of error propagation. We can formulate two equations

$$f(\varepsilon)=\sqrt{2} = \dfrac{2}{3} \dfrac{543}{256} + \varepsilon$$ $$g(\varepsilon)=\sqrt{2} = \dfrac{3}{\dfrac{543}{256} + \varepsilon}$$

The sensitivity of with respect to changes in $$\varepsilon$$ is given by

$$\Delta f \approx \Delta\varepsilon$$ $$\Delta g \approx \dfrac{3}{\left[\dfrac{543}{256} + \varepsilon\right]^2}\Delta \varepsilon.$$

As you can see the first expression scales linearly with $$\Delta \varepsilon$$. The second equation does not scale linearly. Hence, we have different behaviors.

Why should they be the same?

$$\frac {543}{256}$$ was an approximation then the results should be approximations, not exact values.

Let's suppose that $$\frac 32\sqrt2 = \frac 3{\sqrt 2} = w$$ exactly but $$\frac 32\sqrt2 = \frac 3{\sqrt 2}\approx a$$ is an approximation. Now let's suppose the approximation is off by $$\epsilon$$ so that:

$$\frac 32\sqrt2 = \frac 3{\sqrt 2} = a + \epsilon$$.

Okay... so

By $$\frac 32\sqrt 2 = a + \epsilon$$ we get $$\sqrt 2 = (a +\epsilon)\frac 23 = \frac 23a + \frac 23 \epsilon$$. So $$\frac 23a$$ is (apparently) a close approximation of $$\sqrt 2$$.

And by $$\frac 3{\sqrt 2} = a + \epsilon$$ we get $$\sqrt 2 = \frac 3{a+\epsilon}=\frac {3\frac aa + 3\frac {\epsilon}a- 3\frac {\epsilon}a}{a+\epsilon} = \frac 3a - \frac {3\epsilon}{a(a+\epsilon)}$$. So $$\frac 3a$$ is also (apparently) also a close approximation.

However one is off by $$\frac 23\epsilon$$ and the other is off by $$- \frac {3\epsilon}{a(a+\epsilon)}$$. Apparently these are small amounts to be off by. But the are not equal amounts to be off by.

====

FWIW

$$w$$ to about 25 or so decimals is $$2.1213203435596425732025330863145$$ and $$a$$ is $$2.12109375$$, so $$\epsilon =0.0002265935596425732025330863145$$ to 25 or so decimals.

So the error $$\frac 23 \epsilon$$ is about $$1.5106237309504880168872420966667e-4$$ and the error $$-\frac {3\epsilon}{a(a+\epsilon)}$$ is about $$-1.5107851088285175079746363557438e-4$$. Both errors are close. But they aren't the same.