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I am getting confused about a super basic issue.

I have two quantities: $6\pm4$ and $4\pm3$ .

So let's say $x = 6$, $\Delta x = 4$, $y = 4$, $\Delta y = 3$

Now I want to calculate the uncertainty of $z=\frac{x}{y}=1.5$.

If I just take the range of values given by the uncertainties I would get $z=\frac{x}{y}$ somewhere between $\frac{2}{7}\approx0.3$ and $\frac{10}{1}=10$.

Now if I propagate the errors according to $\frac{\Delta z}{z}=\sqrt{(\frac{\Delta x}{x})^2+(\frac{\Delta y}{y})^2}$ then I get $\frac{\Delta z}{z}\approx1.004$ which is an uncertainty of > 100%. I.e. a final result of $z\pm\Delta z\approx1.5\pm1.5$.

This confuses me because firstly, I only predict values up to a maximum of $\approx3$ whereas I predicted values up to $10$ when using the range of possible values. Secondly and more importantly, this uncertainty implies that the ratio of $z=\frac{x}{y}$ can be zero which is odd. Obviously, if I increase the uncertainties of $x$ and $y$ slightly, I can also generate negative nubers of the ratio.

Is there something I am doing wrong? Or a way to deal with this issue? Having zero or negative numbers just doesn't make physical sense at all in the context that I am using these numbers.

Thanks a lot

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A very late answer. But maybe it is useful to someone else..

Your error propagation formula is missing a term. In case of division it should read as: $$\frac{\Delta z}{z}=\sqrt{(\frac{\Delta x}{x})^2+(\frac{\Delta y}{y})^2 - 2\frac{\Delta x}{x}\frac{\Delta y}{y}}.$$

In my opinion it is easier to work with the following Taylor series approximation: in general for $z=z(x,y,...) \longrightarrow \Delta z=\frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y + \dotsb$

If $z =\frac{ x}{ y}$ then $\frac{\partial z}{\partial x} = \frac{1}{y}$ and $\frac{\partial z}{\partial y} = - \frac{x}{y^2}$, so $\Delta z$ simplifies to $\Delta z=\frac{1}{y}\Delta x - \frac{x}{y^2}\Delta y $.

If you now divide this by z: $$\frac{\Delta z}{z}= (\frac{1}{y}\Delta x - \frac{x}{y^2}\Delta y + \dotsb)\frac{y}{x}=\frac{\Delta x}{x} - \frac{\Delta y}{y}. $$

So your result should read $1.5 \pm 0.125$.

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