Q:

How do we know that our definitions/axioms are not contradictory?

Let us assume that we have declared some axioms. Now, we wish to declare a new axiom too. How do we establish that the new axiom is not a consequence of, nor a contradiction to the original set of axioms?
Ditto for definitions. How do we know that the criteria are not contradictory?
For instance,
when creating the set of real numbers axiomatically, our definition says that 1 is not equal to 0, where 1 is the multiplicative identity and 0 is the additive identity.
But, how do we know/prove that 1 not equal to 0 is not a consequence of the field axioms, nor is contradictory to the field axioms?

A:

Let's take an example: the axioms of a group. Now we wish to declare a new axiom: all elements commute. 
How do we establish that this new axiom is not a consequence of the old ones? By constructing a noncommutative group. 
How do we establish that it is not in contradiction to the original ones? By constructing a commutative group.
Here are a couple of other examples, using exactly the same proof scheme. 
The parallel postulate is neither a consequence of nor a contradiction to the other axioms of Euclidean geometry: construct the Cartesian coordinate plane with distances and angles etc., to show that it is not in contradiction; and construct the hyperbolic plane to show that it is not a consequence.
The axiom of choice AC is neither a consequence of nor a contradiction to the other axioms ZF of set theory: construct the set theoretic model of constructible numbers to show that AC is not in contradiction to ZF; use forcing to construct models which show that AC is not a consequence of ZF.

A:

In a logical approach:
Assume we have a theory $T$ and its set of axioms $\Gamma$, such that $T=\{\varphi|\Gamma \vdash \varphi\}$ (a theory is closed under logical consequence). Now, for some reason, we want to postulate a new axiom to $T$, call it $\psi$. Clearly, we have a new axiom set $\Gamma'=\Gamma \cup \{\psi\}$ and so a new theory $T'=\{\varphi|\Gamma' \vdash \varphi\}$.
Your question is how to proof that:

$T \neq T'$
$T \cup \{\psi\} \nvdash \bot$

is the case, assuming that $T$ is consistent.
Then:

We need to establish an independence proof of $\psi$ in $T$, that is, we need to show that $T \nvdash \psi$. Mathematicians usually do this using models (see the completeness theorem), and the matter turns out to find an interpretation $\mathcal{M}$ of $T$ where $\psi$ is false. This shows that $T \nvdash \psi$ and that $T \subset T'$, since $T' \vdash \psi$.
The matter is how to give a consistency proof of $T \cup \{\psi\}$. In other words, this means that adding $\psi$ is consistent with the axiom set of $T$. Since $T$ is consistent, it suffices to show that $T \nvdash \neg\psi$. In this case, we need to show that there is a model $\mathcal{M}$ of $T$ where $\psi$ is true.

