Is $dS=\frac{\delta Q_{irev}}{T}$ true for non-reversible processes? The German Wikipedia reads

Das Differential $\mathrm {d} S$  ist nach Clausius bei reversiblen Vorgängen zwischen Zuständen im Gleichgewicht das Verhältnis von übertragener Wärme $\delta Q_{\mathrm {rev} }$ und absoluter Temperatur $T$: $dS=\frac{Q_{\mathrm {rev} }}{T}$

Which translates to

According to Clausius the differential $\mathrm {d} S$  for reversible processes between equilibirum states is the ratio between transmitted heat $\delta Q_{\mathrm {rev} }$ and absolute temperature $T$: $dS=\frac{Q_{\mathrm {rev} }}{T}$

This formulation seems confusing to me. Why do we need reversibility? I do not see why this shouldn't be true for quasi-static irreversible processes. We start at a state of entropy $S_1$ and by some process we reach $S_2$. As the entropy by axiom is path-independent it shouldn't matter weather the path is reversible or not.
Addendum: Many people stated in the comments that one can use a reversible process starting and resulting in the same equilibrium state, as the irreversible one. While this is true and an important concept, my question was aimed at the actual heat $\delta Q_{irev}$ that is transferred to the system during a irreversible process. 
Related The actual definition of entropy
 A: 
Why do we need reversibility? I do not see why this shouldn't be true
  for quasi-static irreversible processes.

Although the definition is in terms of a reversible transfer of heat, you are correct that it is not limited to a reversible process, i.e., it applies to an irreversible process as well. Entropy is a state function or property, like internal energy. That means the difference in entropy between two equilibrium states is independent of the path or process between the states. 
So if you have an irreversible process taking you between two states you can determine the entropy change of the system by assuming any convenient reversible process between the states. That will give you the entropy change for the system for the irreversible process as well since entropy is a state function.
However, if the process is irreversible, entropy is generated by the system. In order to return the system to its original state (perform a cycle) the entropy generated will need to be transferred to the surroundings making the total entropy change (system + surroundings) >0 for a complete cycle. For a reversible cycle the overall entropy change = 0.
Hope this helps.
A: One counterexample is a quasi static irreversible adiabatic free expansion. Here d$S>0$ and d$Q=0$, so the equality is not valid for this irreversible process.
A: Just for a comment;
The discussion here seems to confuse two (three) different types of "entropy".


*

*If you say that "entropy is a state quantity" this is what is called "exchanged entropy". This is uniquely determined depend on the state (U,V,N), so let's represent this as a mathematical multivariable function $S_{e}(U,V,N)$.

*If you say that "entropy increases with decreasing irreversible processes", this is called "generated entropy". Let's express this by the symbol $S_{g}$.

*We define the  $\Delta S_{tot}$ as follows. The  $\Delta S_{tot}$ represents the entropy change in the entire system via the reaction between  the reaction from $(U_1,V_1,N_1)$ to  $(U_2,V_2,N_2)$ ;
$$\Delta S_{tot}:=S_{e}(U_2,V_2,N_2)-S_{e}(U_1,V_1,N_1)+S_{g}$$

For more details for the exchanged/generated entropies plesae
this book might be helpful.
Note that since $S_{g}$ is not a state quantity, $S_{tot}$ is not a state quantity either.
This Q&A also includes other confusion.
The concept of "whether this reaction can be written as a curve in U-V-N space or not" and the concept of "whether this reaction is reversible or not" are different concepts. In this sense, the reaction paths can be classified from the two different view point; "writable/unwritable" and "reversible/irreversible".
From the "writable/unwritable" viewpoint, the reaction paths can be classified into two types;


*

*Reaction that cannot be written as smooth curves in the state space(U-V-N space): e.g., "adiabatic free expansion

*A reaction that can be written as a smooth curve in the state space: the quasi-static process is pseudo-this.


The viewpoint of "reversible/irreversible", is too confused, so, I will omit the definition of this. However, there are four kinds of combinations:

*

*writable and reversible,

*writable and irreversible,

*unwritable and reversible, and

*unwritable and irreversible.

Probably the third one is an abstract nonsense (Such an example probably does not exist.)　
Furthermore, "adiabatic free expansion" has　also been confused with "quasi static adiabatic expansion. " In order to be "quasi static", the piston must be controlled; move a little bit, applying the brakes. This is no longer free expansion. In the process of braking the piston, an external force is applied to the system.
A: Clausius' theory is about heat exchange with quasistatic volume change. When the volume is changed non quasistatically (as in free expansion or when some irreversible work is done), the question takes a different meaning. In Clausius' approach the only changes in entropy are caused by heat exchange, not irreversible volume change.
Orginally, Clausius' point about irreversibility is the following. When a system exhanges heat with the surroundings (a heat bath), we have:

*

*the heat transfer is reversible if and only if $T_{surr}=T_{system}$

*then, $\Delta S_{system} = \frac{Q}{T_{surr}}$

*otherwise $\Delta S_{system} > \frac{Q}{T_{surr}}$
Important: "reversible/irreversible" here applies to the heat exchange, meaning to "system + surroundings". Seeing the process as reversible/irreversible for the system only does not make sense and leads to confusions.
Writing just $T$ is ambigous. Is it the temperature of the system or the temperature the surroundings? The fact is, when heat is not transfered reversibly, the temperature of the system usually does not even exist, since a gradient appears inside the system and it does not have a uniform temperature. Assuming the gradient spreads over a small volume (negligible energy), we may say the system still has constant temperature almost everywhere. Then we have $\Delta S_{system} = \frac{Q}{T_{system}}$.
As a conclusion, $T$ is meaningful for reversible processes, and then it is both the temperature of the system and the surroundings. Otherwise, $T_{system}$ does not exist during the process and $\Delta S_{system} > \frac{Q}{T_{surr}}$. When the temperature is near uniform inside the system, then $\Delta S_{system} = \frac{Q}{T_{system}}$, even for an irreversible process.
