11,069 reputation
11230
bio website math.brown.edu/~lubinj
location Minnesota
age 77
visits member for 2 years, 6 months
seen 2 hours ago

Aged mathematician


Apr
3
comment Galois Theory: Finding the discriminant of a polynomial
You know that $s_1$ is the sum of the roots, and $s_3$ is their product, right?
Apr
3
answered Lost on Galois Group over $\mathbb{Q}(i)$ Question.
Apr
2
comment Can a transcendental number be an infimum of a set of rationals?
No, if you had more examples under your belt, you would have a different feeling. How about the set of rationals between $1$ and $2$ for which the cosine is negative?
Apr
1
comment Union of field extensions over Q
For my money, this proof suffices, as long as you understand that the union actually is a field. Is the sum (product) of any two elements an element of the field? What if one is in $K_6$ and the other is in $K_7$, where $K_n=\mathbb Q(2^{1/n})$ ?
Apr
1
comment Does a basis of three vectors always span $R^3$
Make sure the three vectors you’re picturing point in three entirely different directions.
Apr
1
comment Proof that any square is of the form $3k$ or $3k+1$
Better to verify it for $n=0,1,2$ and then do three inductions, going from $n$ to $n+3$ in each.
Apr
1
comment Why is cosine a sine function with offset pi/2?
Do you know the addition and subtraction formulas for sine and cosine? Alternatively, following the suggestion of @user88595, can you use reflection in the line $y=x$ to relate the sine of $\theta$ to the cosine of $\pi/2-\theta$?
Mar
31
comment $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
@angryavian, here’s what I (unsatisfactorily) say: it’s, quote, well known, unquote, that $\{1,2^{1/3},4^{1/3}\}$ is an integral basis for $\mathbb Q(2^{1/3})$. I know I’ve actually verified this with pencil and paper myself, and as I recall, it’s a slow slog.
Mar
29
comment Remainder and Quotient in Mod 7
The two remainders you correctly offer are one and the same. Your quotient is right, too.
Mar
28
comment Maclaurin Series Complex Numbers
Well, you know that $1/(1+z^2)=1-z^2+z^4-z^6+z^8-\cdots$.
Mar
28
comment $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
Well, I cubed the expression out, and looking at the constant term, I saw that $a$ had to be odd; at the next term that $b$ had to be even; and at the third that $c$ also had to be even. Then it seemed to me that you could prove inductively that $b$ and $c$ were divisible by every power of $2$. So they had to be zero, showing that $a^3=5$. Of course, I was assuming that $a$, $b$, and $c$ were integers! But a bit of thought ought to get you through that. Anyhow, I hafta go to bed.
Mar
28
answered $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
Mar
28
comment Find the angle of intersection of circles $x^2+y^2-6x+4=0 \ \&\ x^2+y^2-2x-2y-8=0$
You should tell us the point of intersection. Ordinarily, by the way, an instructor would be looking for an answer of the form $\arctan(\text{some rational number})$
Mar
28
comment If $\lim f(x)$ and $\lim g(x)$ do not exist, can the $\lim [f(x)+g(x)]$ exist?
You should have a few examples in your bag of tricks. Do you have a function that has no limit at (for instance) $a=0$? And a function that does? Call the first one $g$ and the second one $f$. What about $f+g$? What about $f-g$?
Mar
27
comment Rigorous Proof: Circle cannot be embedded into the the real line!
Well, the circle is disconnected by the removal of two points. So the argument has to be changed. Instead, the image of the circle in the line is both connected and compact, thus a closed bounded interval, which can (by removing the endpoints) remain connected after the removal of two points. But the circle is always disconnected after the removal of two points.
Mar
27
answered Rigorous Proof: Circle cannot be embedded into the the real line!
Mar
26
comment Relation between divisibility of polynomials in different rings, $h | f$ in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$
Your bulleted items don’t mention $f$ nor $g$. Could you please edit?
Mar
26
awarded  field-theory
Mar
25
answered What does it take to get a job at a top 50 math program in the U.S.?
Mar
25
comment Show that if $a$,$b$,$c$, and $d$ are integers, where $a\neq 0$, such that $a|c$ and $b|d$, then $ab|cd$.
Yes, once you write the definition down and what it means in the hypothesized cases, you see your way to the conclusion.