Why is Decisional Diffie–Hellman assumption harder than the Computational Diffie–Hellman assumption in cryptography?
There are groups that allow a 'pairing' operation; that is, an efficiently computable function $e$ with the property:

$$e(x, y^a) = e(x^a, y)$$

for all group members $x, y$ and integers $a$.  Actually, the pairing operation maps the pair into another group, with the additional property $e(x^a, y^b) = e(x, y)^{ab}$; that's not important here (but is important for other uses of the pairing function).  We also assume that $e$ is nontrivial, that $e(x, y)$ doesn't uniformly map to the identity group member.

In any case, if we were given four values $g, g^a, g^b, g^c$, this function would allow us to check whether:

$$e(g, g^c) \stackrel{?}= e(g^a, g^b)$$

If this equation doesn't hold, then we know that $g^{ab} \ne g^c$ (and if we make some assumptions on $e$, then we can actually get equivalence).

However, this $e$ function doesn't give us any obvious way to recover the value $g^{ab}$, given $g, g^a, g^b$