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Let \(f(x) = \sin(x^3)\), suppose it is periodic, and let \(T\) be a period of \(f\). Since \(f\) is differentiable it follows that \(T\) is a period of \(f'\) as well. Observe that \(f'(x) = 3x^2 \cos(x^3)\) is continuous and so is bounded on \([0,T]\). Since \(T\) is a period of \(f'\), it follows...

Suppose that \(A \neq 0\), say \(A_{ij} \neq 0\) for some \(i,j\). Let \(P\) be any permutation matrix with \(P_{ij}=1\) and let \(Q\) be the matrix obtained from \(P\) by changing its \(ij\)-entry to \(0\). Finally let \(X = xQ\) where \(x \in \mathbb{C}\). We have that \(\det(X) = 0\) and that \(\...

Hi Alkes. I get that the characteristic is equal to \(2\) in exactly the same way as you do. In fact, for \(x \neq 0\) we can be quicker. Since we are in a field we know that \(2x=0\) and \(x\neq 0\) imply that \(2=0\). (We can take this for granted, but the proof is the one that you give.)

Nice question! Suppose \(\varphi: (R,+) \to (R \setminus \{0\},\cdot)\) is an isomorphism. We must have \(\varphi(0)=1\). There is an \(x \in R\) with \(\varphi(x)=-1\). Then \(\varphi(2x) = \varphi(x)^2 = 1 = \varphi(0)\). So we must have \(2x = 0\) and so either \(x=0\) or \(2_R = 0\). In both cas...

If \(R\) is a division ring, given \(a\neq 1\), consider the non-zero element \(a-1\) of \(R\). It has a multiplicative inverse \(x\). So \((a-1)x=1\). Taking \(b=x+1\), we get \(1 = (a-1)x = (a-1)(b-1) = ab-a-b+1\) giving \(ab=a+b\) as required. Conversely, if \(R\) satisfies the property, given \(...

We observe that complex conjugation is a field automorphism of \(\mathbb{K}\). If \(f(x)\) has a non-real root, then \(\mathbb{K}\) has a non-real element and therefore this automorphism has order \(2\). Let \(\mathbb{L}\) be the fixed field of the automorphism. (I.e. \(\mathbb{L}=\mathbb{K} \cap \m...

I believe there is a small problem towards the end of the solution. It is not necessary that \(\rho^p \in K\). What we can say is that there is an element \(\alpha \in K(\rho)\) with \(\alpha \notin K\) such that \(\alpha^p \in K\). From this we can derive a contradiction as follows: Since \(\alpha ...

No there is not. Suppose \( \phi: D_4 \to \mathbb{Z}_4\) is a homomorphism. Then \[ 3\phi(s) = \phi(s^3) = \phi(tst) = \phi(t) + \phi(s) + \phi(t) \] and so \(2\phi(s) = 2\phi(t).\) But since \(2\phi(s) = 2\phi(t) = \phi(t^2) = 0\), then \(\phi(s),\phi(t) \in \{0,2\}\). So \(\phi(D_4) \subseteq \{0,...

We observe that if \(x = a_1\) then the first row of \(A\) is identical to the second row and so in this case the determinant is 0. This implies that \(x-a_1\) is a factor of the determinant. Similarly, \(x-a_2\) is also a factor since in the case that \(x=a_2\) the second row of \(A\) is identical ...

Suppose that $A \neq 0$, say $A_{ij} \neq 0$ for some $i,j$. Let $P$ be any permutation matrix with $P_{ij}=1$ and let $Q$ be the matrix obtained from $P$ by changing its $ij$-entry to $0$. Finally let $X = xQ$ where $x \in \mathbb{C}$. We have that $\det(X) = 0$ and that $\det(X) = \det(A+X) - \det...