Your question touches on many aspects of mathematics and physics at once - here is how I see it:

In (linear) algebra, the exterior algebra arises (form one viewpoint) solely through the need to define the determinant in a non-basis-dependent way, since it is, up to normalization, simply the sole basis element of the top exterior power. Since the determinant for matrices who possess two linearly dependent columns is zero, $x \wedge x = 0$ is the way to enforce this, and over fields with characteristic not 2, this implies antisymmetry. The actual way we construct this exterior power is by dividing the ideal generated by $I := \{x\otimes x | x \in V\}$ out of the tensor algebra over a vector space $V$, so we can see the exterior power as consisting of antisymmetric tensors, although sometimes one should remember that it really is a quotient of the tensors. In this way, antisymmetry of the exterior power arises *naturally*.

Now, in differential geometry, the differential forms arise through the cochain complex of the de Rham cohomology, which is just the complex of $p$-forms with the exterior derivative as the coboundary. You ask: "Why do we take the degrees of the exterior power of $T_p M$ and not the degrees of the tensor algebra as defining these forms?" The answer is simple: On mere tensors, $\mathrm{d}^2 = 0$ would not hold, thus the degrees of the tensor algebra do not define a cochain complex, thus they do not generate a cohomology. If you ask why we want a cochain complex/cohomology, it is because, by virtue of Eilenberg-Steenrod, every (co)homology is intrinsically related to the geometric intuiton of loops, surfaces and hypersurfaces on a manifold.

Fermionic and bosonic functions/objects are a wholly different story: They arise through different representations of the Lorentz group $\mathrm{SO}(1,3)$, or rather, its double cover $\mathrm{Spin}(1,3)$. Since we want Lorentz invariance, the Lorentz group must have a way to act on our fields/objects. It can only do so if they are in a representation of our symmetry group, and when we construct reps, we generally do so by representing only the infinitesimal generators of the symmetries, which lie in the Lie algebra - but this is the same for the Lorentz group and its double cover, and representing the algebra will always induce reps of the universal cover, so we are stuck with the reps of $\mathrm{Spin}(1,3)$, and this has representations with "spin" eigenvalue $j \in \mathbb{N}$ and $j \in \mathbb{N}+\frac{1}{2}$. Fields transforming in one of the former reps are bosonic, fields transforming in one of the latter are fermionic.

It's almost the defition. A tensor $T_{ab}$ of rank $2$ is symmetric if, and only if, $T_{ab}=T_{ba}$, and antisymmetric if, and only if, $T_{ab}=-T_{ba}$. So from this definition you can easily check that this decomposition indeed yields a symmetric and antisymmetric part.

**Edit:** Let $S_{bc}=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)$. Then
$$S_{cb}=\dfrac{1}{2}\left(A_{cb}+A_{bc}\right)=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)=S_{bc},$$
so, $S_{bc}$ is symmetric. On the same way, if $T_{bc}=\dfrac{1}{2}\left(A_{bc}-A_{cb}\right)$, we have
$$T_{cb}=\dfrac{1}{2}\left(A_{cb}-A_{bc}\right)=-\dfrac{1}{2}\left(A_{bc}-A_{cb}\right)=-T_{bc},$$
and $T_{bc}$ is antisymmetric.

## Best Answer

I) Many English words come in both a Greek and a Latin version. The prefix

anti-is from Greek and the prefixskew-is from French.Most authors would define an

anti-symmetricand askew-symmetric(possibly higher-order) tensor as precisely the same thing.II) However, in the context of supernumber-valued tensors, some authors define a second-order anti-symmetric tensor/matrix as

$$\tag{A} A_{ab}=(-1)^{(|a|+1)(|b|+1)}A_{ba},$$

while a second-order skew-symmetric tensor/matrix obeys

$$\tag{S} S_{ab}=-(-1)^{|a||b|}S_{ba},$$

cf. Ref. 1. Here $|a|$ denotes the Grassmann-parity of the coordinate index $a$.

References:

Seminar on supersymmetry. Vol. 1. Algebra and Calculus,2006.