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We can construct a division algebra without multiplicative inverses?

In the article the octonions (http://math.ucr.edu/home/baez/octonions/node2.html), this is possible by taking the quaternions and modifying the product slightly, setting $i^2 = -1 + \epsilon j$ for some small nonzero real number $\epsilon$ while leaving the rest of the multiplication table unchanged.

Gabriel Fanini
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No, that is even implicitly required in the definition you're citing:

An algebra $A$ is a division algebra if given $a,b \in A$ with $ab = 0$, then either $a = 0$ or $b = 0$. Equivalently, $A$ is a division algebra if the operations of left and right multiplication by any nonzero element are invertible.

Earlier:

For us a vector space will always be a finite-dimensional module over the field of real numbers. An algebra $A$ will be a vector space that is equipped with a bilinear map $m : A \times A \to A$ called 'multiplication' and a nonzero element $1 \in A$ called the 'unit' ...

If $a$ is nonzero, then left multiplication by $a$ is an injective endomorphism of a finite dimensional vector space, and hence it is also surjective. This means there exists a $b$ such that $ab=1$.

The whole point of a division algebra is that you can divide...

rschwieb
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    When Baez says $i$ fails to have a multiplicative inverse in the modified quaternions, he means it fails to have a two-sided inverse (the left and right inverses of $i$ are different, which is possible in a nonassociative algebra). It is a separate exercise to check that all nonzero elements have both left and right multiplicative inverses (even if they aren't always equal). – anon Apr 06 '18 at 23:40