transfer function - Why more poles than zeroes?

I read that an "improper system" "has more zeros than poles; it is not causal, cannot be implemented, has a strictly proper inverse and has infinite high-frequency gain."

Does causality fail due to instant signal change? I try to make up a recurrence, where $x_n$ depends on both $x_{n-1}$ as well as $x_{n+1}$ but they all can be interpreted as $x_{n+1}$ dependence on $x_{n}$ and $x_{n-1}$ and, thus, are causal.

Is it related with solving recurrent relations saying that P(x) in generating function $f(x) = f_0 + f_1 x + f_2 x^2 + \cdots = P(x)/R(x) \ \ \ \ \ $ must have degree less than R(x) as Miguel A. Lerma says in Generating function of Linear Homogenous Recurrence. Here, $R(x) = 1 + r_1 x^1 + r_2 x^2 + \cdots$ is a reciprocal of the characteristic polynomial of our recurrence $x_n + c_1 x_{n+1} + \cdots = 0.$ This indeed seems to happen whenever which R(x) I try.

On the other hand, https://ccrma.stanford.edu/~jos/fp/Existence_Z_Transform.html seem to identify causality with pole values (if your pole > 1 then series does not converge, which means anticausality) rather than their number and others say that causality is $x_n = 0$ for all $n < 0.$ I do not see how this is related with poles/zeros ratio and my guess that poles > zeroes = causality may be wrong. I have asked to relate various kinds of causality here.

Answer

I think the system is always non-causal if it has more zeros than poles. @Hilmar, your implementation might be possible digitally but if you see a transfer function like this: $H(z) = \frac{(z-1)(z-2)}{(z-3)}$, taking the inverse Z-Transform, $y(n-1)-3y(n-2) = x(n)-3x(n-1)+2x(n-2)$ is clearly non-causal and cannot be implemented real-time(of course you can store the signals and do processing offline, but real time implementation is impossible). So I guess Non-Causality is the answer, apart from the infinite high-frequency gain.

Update by Val Ok, I understand now. In wikipedia, we see that $$a_0 y_n + a_1 y_{n-1} + a_2 y_{n-2} + \cdots = b_0 + b_1 x_{n-1} + b_2 x_{n-2} + \cdots$$ is identical to $$H(z) = {b_0 + b_1 z^{-1} + b_2 z^{-2} + \cdots \over a_0 z^{0} + a_1 z^{-1} +\cdots}.$$ I have finally noted that there are no positive powers. That is, $a_0$ corresponds to the current output, $z^0 = y[n]$, and $\cdots = a_{-2} = a_{-1} = \mathbf{0} = b_{-1} = b_{-2} = \cdots$. If we had more zeroes then we have more power in the nominator and, after normalizing $H(z)$ so that $a_0$ has $z^0$, we still have a positive power of z in the nominator, which means that output depends on future input.

In the simplest case, $Y(z) = zX(z)-x_0$ corresponds to $y[i] = x[i+1]$, that is output depends on future input (we can also drop $x_0$ assuming our sequences are causal, i.e. $0 = y_{-1} = x_0$ aka $z^{-1}Y = X$ or $Y = zX$). It is curious that causal is an example of non-causal. Could anybody clarify it further?

It also worth noting that $(1-az^{-1})(1-bz^{-1})$ is not a zero of order two. It has equal number of zeroes and poles, since it is equal to $(z-a)(z-b)\over z^2.$