Show that \(\mathbb{R}^{2n+1}\) is not a division algebra over \(\mathbb{R}\) if \(n > 0\) by considering how the determinant of the linear map \(x \mapsto ax\) given by the multiplication in a division algebra structure would vary as \(a\) moves along a path in \(\mathbb{R}^{2n+1} - \{0\}\) joining two antipodal points.
Solution
Assume for contradiction that \(A=\mathbb R^{2n+1}\), with \(n>0\), has the structure of a real division algebra. For \(a\in A\) let \[ L_a:A\to A,\qquad L_a(x)=ax. \]
1. If \(a\ne0\), then \(L_a\) is invertible, hence\[ \det L_a\ne0. \]
In a division algebra the equation \(ax=b\) is solvable for every \(b\) whenever \(a\ne0\), so \(L_a\) is surjective. Since \(A\) is finite-dimensional, \(L_a\) is therefore an isomorphism.
2. Choose \(a\ne0\). Since \(2n+1\ge3\), there is a path\[ \gamma:[0,1]\to A-\{0\} \]with \(\gamma(0)=a\) and \(\gamma(1)=-a\).
The punctured Euclidean space \(\mathbb R^{2n+1}-\{0\}\) is path-connected when \(2n+1>1\); for instance join antipodal points by a semicircle on the sphere of radius \(\|a\|\).
3. The function\[ t\longmapsto \det L_{\gamma(t)} \]is continuous and never zero, but its endpoint values have opposite signs.
Bilinearity of multiplication implies that the matrix entries of \(L_a\) depend linearly, hence continuously, on \(a\). Thus the determinant depends continuously on \(a\). By 1 it never vanishes along \(\gamma\). On the other hand \[ L_{-a}=-L_a, \] so, because the dimension \(2n+1\) is odd, \[ \det L_{-a}=(-1)^{2n+1}\det L_a=-\det L_a. \]
4. This is impossible.
A continuous nonzero real-valued function on the connected interval \([0,1]\) has constant sign. The endpoint calculation in 3 contradicts this.
Hence \[ \boxed{\mathbb R^{2n+1}\text{ admits no real division-algebra structure for }n>0.} \]