Why is a quaternion a + bi + cj + dk not equal to a + (b+c+d)i

Question

i² = -1

j² = -1

k² = -1

But

a + bi + cj + dk is not equal to a + (b+c+d)i

why not ?

Answer 1

Because $i, j, k$ aren't equal. In the quaternions, $-1$ has even more square roots than it does in the complex numbers.

Answer 2

$$\forall a,b,c,d,a',b',c',d' \in \Bbb R: a + bi +cj +dk = a'+ b'i+c'j+d'k \iff a=a', b=b', c=c', d=d'$$ just as if we'd have written a $4$-tuple $(a,b,c,d)$ for the quaternion. The coefficients uniquely determine the number.

In particular $i = 0+1i + 0j + 0k \neq j = 0 + 0i + 1j + 0k$ etc. The roots of $-1$ are $i,-i,j,-j,k,-k$ already showing the quaternions are not a field (already obvious from non-commutativity: $ij=k, ji=-k$ etc.) as in a field a quadratic like $x^2+1$ can have at most $2$ roots, not $6$.