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.

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

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.

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.