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bio website math.brown.edu/~lubinj
location Minnesota
age 77
visits member for 2 years, 6 months
seen 5 hours ago

Aged mathematician


Apr
5
comment Continuous bijective function between $\Bbb{R}$ and $[0,1]$?
To be clear, you’re not asking for the inverse function to be continuous, right?
Apr
5
comment Is (x,0) the only solution?
Sorry, I was on a plane for a few hours. As you have stated the problem, the “square” issue is beside the point. Clear your mind of cases when $n$ is a square and answer the question fully: which numbers are the difference of two squares? If you’re having difficulty with this, write down a list of the ten first squares, and see which numbers are the difference of two things on your list. I keep harping on this, I know, but knowledge proceeds from examples. Your problem is that you haven’t been thinking of examples. (AND MAKE SURE YOUR LIST OF SQUARES INCLUDES ZERO.)
Apr
4
comment Is (x,0) the only solution?
But the problem itself is asking which numbers are difference of two squares, right?
Apr
4
comment Is (x,0) the only solution?
Nothing in the problem seems to ask for $n$ to be a square.
Apr
4
comment Invalid subtraction when solving system of equations?
Check them all! The job is not done until you do. Sometimes square-rooting will give you false solutions...
Apr
4
comment Invalid subtraction when solving system of equations?
Here in the US, we don’t make it clear to students what they’re doing when they “solve” an equation. That’s really the uniqueness part of the story: here, if there is a solution, it’s either $(1/2,1/2))$ or $(0,0)$. But the existence part is what we call “checking”: to see whether the possibilities you found really do satisfy the equation(s). Here, OP has done the checking and found that one of the apparent possibilities is in fact not a solution. In teaching, we haven’t been stressing strongly enough that checking is an essential part of the process.
Apr
3
comment Quotient field of a localization
If by “quotient field” you mean the total ring of fractions, it is true, and you show it simply by writing down what the elements of each turn out to be.
Apr
3
comment When taking subsequent derivatives, why are units squared?
Precisely. Remember that the definition of derivative always involves a division. If you’re differentiating twice with respect to time, there’ll be a “per second$^2$” in the units.
Apr
3
comment Galois Theory: Finding the discriminant of a polynomial
(my last comment) You can use your formula, just multiply it all out, then start subtracting off monomials in the $s$’s, starting with $27s_3^2$, and it all comes together. It is a computational pain, though, but a good exercise.
Apr
3
comment Galois Theory: Finding the discriminant of a polynomial
Product, you mean, of course. It might help if you also make use of the fact that the discriminant is the product of $F'(\rho)$, where $\rho$ runs through the roots. This equivalence is not hard to check.
Apr
3
comment Galois Theory: Finding the discriminant of a polynomial
It would depend on which definition of the discriminant you’re using.
Apr
3
comment $\mathbb Q[x]/\langle x^2+1\rangle$ is a splitting field of $x^2+1$ over $\mathbb Q.$
Doesn’t it have all the roots of your polynomial? And isn’t it the case that no smaller field has the roots? That’s my idea of splitting field.
Apr
3
comment Property of isomorphic normal subgroups
As it is written, this question must be interpreted the way you do. Unfortunately when our understanding is murky, we often can’t even express our questions clearly.
Apr
3
comment Lost on Galois Group over $\mathbb{Q}(i)$ Question.
@HenrySwanson, I’m not sure which “primitive roots” you’re talking about here. If the roots of unity, the automorphisms are those that take $\zeta$ to an odd power; if the roots of $X^8-i$, the automorphisms are those that take $\zeta$ to an $m$m-th power, $m\equiv1\pmod4$.
Apr
3
comment Galois Theory: Finding the discriminant of a polynomial
If I read this comment right, you should get the same discriminant.
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.