# How can this binomial expansion result in two different approximations of root 2?

I have been working on a problem on approximating $$\sqrt{2}$$ using the first three terms of the following binomial expansion, and a substitution of $$x = -\frac{1}{10}$$ :

$$(4 - 5x)^.5 = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$

After substituting, I got to the stage:

$$\frac{3}{\sqrt{2}}= \frac{543}{256}$$

Now what is bizarre, is that if I solve this equation for $$\sqrt{2}$$, there are two routes I could take, but they give slightly different answers.

Route 1:

Reciprocate both sides, and multiply both sides by 3 to get:

$$\sqrt{2}= \frac{256}{181}$$

Route 2:

Rationalize the left hand side to make the equation:

$$\frac{3\sqrt{2}}{2}= \frac{543}{256}$$

Divide both sides by $$\frac{3}{2}$$:

$$\sqrt{2}= \frac{181}{128}$$

So we end up with two approximations of $$\sqrt{2}$$: $$\frac{256}{181}$$ and $$\frac{181}{128}$$

I am really struggling to understand how this has happened? Why is the same equation leading to two different solutions?

• You did not actually start with an equation, but with an approximation. – Peter May 16 '19 at 12:04
• One weird thing that I notice is in the part "rationalize the left hand side". You start with $$\sqrt{2} = \frac{256}{181}$$ for which the inverse is $$\frac{1}{\sqrt{2}} = \frac{181}{256}$$ Multiply both sides by $3$ to get $$\frac{3}{\sqrt{2}} = \frac{543}{256}$$ And now, you have used the relation $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, but notice that this is not equivalent to the definition of $\sqrt{2}$ in the beginning. I think this is the source of the error. – Matti P. May 16 '19 at 12:08
• Neither is exact: In fact $\sqrt{2}$ is closer to $\dfrac{181.019336}{128}$ and $\dfrac{256}{181.019336}$ since $181.019336 \approx 128 \sqrt{2} = \dfrac{256}{\sqrt2}$ with your approximations rounding ${181.019336}$ to $181$ in both cases – Henry May 16 '19 at 12:08

If we denote the approximation with $$\ c\$$, we have two equations that we can solve for $$\ x\$$ :

• $$\ \frac{3}{x}=c\$$ giving $$\ x=\frac{3}{c}\$$
• $$\ \frac{3x}{2}=c\$$ giving $$\ x=\frac{2c}{3}\$$

If $$\ c\$$ were exactly $$\ \sqrt{2}\$$, both solutions would coincide. But $$\ c\$$ is only an approximation, hence the values cannot coincide excactly. Their product , however , is exactly $$\ 2\$$ , so one approximation is too small and the other too large.

Let $$f(x) = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$.

Let $$c$$ denote the value $$f(-\frac{1}{10}) = \frac{543}{256}$$.

We can write

$$\tag 1 \frac{3}{\sqrt 2} = c + \varepsilon$$

If we take the OP's route 1 (but keeping $$\varepsilon$$), then

$$\tag 2 \sqrt 2 = \frac{3}{c+\varepsilon}$$

We can check that the 'route 1 approximation', $$\frac{3}{c}$$, is strictly greater than $$\sqrt 2$$. With $$\varepsilon$$ playing a part in the denominator of the rhs of $$\text{(2)}$$, to 'fix things up' it must be true that $$\varepsilon \gt 0$$.

If we take the OP's route 2 (but keeping $$\varepsilon$$), then

$$\tag 3 \sqrt 2 = \frac{2}{3}\,(c + \varepsilon)$$

Since $$\varepsilon \gt 0$$, the 'route 2 approximation' $$\frac{2c}{3}$$ is strictly less than $$\sqrt 2$$.