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age 19
visits member for 2 years, 3 months
seen Oct 20 at 23:28

I am a high school student interested in mathematics.


Apr
9
comment Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24)
Since $b^2$ is a perfect square, we have that $b^2 = 1,4,9,16$ modulo $24$. All of these, besides $1$, require that $2 | b$ or $3 | b$.
Mar
9
comment Is this fraction even possible to put into partial fractions?
@Wolff Think inverse trigonometric functions.
Feb
6
comment Real Analysis Question concerning existence of curve and derivative?
Suppose $f(0) \ne 0$. Use continuity of $f$ to obtain a contradiction.
Jan
31
comment % of my equivalent human life hunting a fly for 5 minutes before I kill it, if its lived for 3 days and myself for 40 years
House flies usually live for 15 to 30 days once hatched from pupal stage, though you may very well have been chasing a 3-day-old fly.
Jan
24
comment Does an element of a group to the 0th power equal the identity?
In other words, the assignment of $x^0 = 1$ plays nice with $x^ax^b = x^{ab}$ for all $a$ and $b$, a desirable property.
Jan
19
comment Does this equation have positive integer solutions?
In other words, combining any two Pythagorean triples gives a solution.
Jan
5
comment How do I show using math symbols “get quotient without remainder”
Try $\lfloor 10/3 \rfloor=3$. This is the "floor" operation and gets rid of any decimal.
Jan
2
comment How find this integral
Your bounds are for $t$, but you treated them as if they were for $\theta$. Notice that $\theta = \sin^{-1} t$, so you can easily switch back.
Dec
27
comment Don't be Greedier
If a player removes one pebble when there is an odd number of pebbles remaining, that player wins.
Dec
27
comment Don't be Greedier
What have you tried?
Dec
19
comment How to justify what solving method to use
Every quadratic can be done by completing the square -- it is the equivalent of the quadratic formula I believe.
Dec
17
comment The intuitive understanding of $\sqrt{x}$ to model “inversely proportional / inverse square”?
I am confused as to why my answer is being voted down. Is something I said incorrect or misleading?
Dec
16
comment Find derivative of $f(x)=(x+2)^{(x-1)}$
Others might be taking the natural logarithm of both sides and then using implicit differentiation (chain rule, really). As in: $f'(x) / f(x) = D_x ((x-1)\ln(x+2))$.
Dec
16
comment Show that the sum of the series
I deleted my answer since my initial premise was faulty.
Dec
16
comment How many 6-letter words that have either exactly 2 vowels or 4 vowels are there? (all lower case)
@mharris7190 If your follow-up is not already answered, I have answered above. For others, the follow-up was about viewing the problem from the perspective of putting distinguishable balls into distinguishable boxes without exclusion--a very good question.
Dec
16
comment How many 6-letter words that have either exactly 2 vowels or 4 vowels are there? (all lower case)
Take it from the slot's perspective. It must choose one of the $5$ vowels. Since there are $2$ slots, there are $2$ factors of $5$, or $5^2$. The difference from ball-in-box problem is that multiple balls can go into each box, but slots cannot have multiple vowels. We can reduce it to a ball-in-box problem by saying the slots are the balls and the vowels are the boxes. Each slot picks a vowel to "go into". Notice that each vowel can "have" more than one slot, since there can be repeats--just like the ball-in-box problem! But now since the roles are reversed, the answer is $5^2$, not $2^5$.
Dec
16
comment How many ways to $22$ balls in $5$ boxes problem
If the boxes are identical, the answer is the number of partitions of $22$.
Dec
16
comment Real life examples of commutative but non-associative operations
This depicts $1 + 2 \ne 2 + 1$, a non-commutative system, not a non-associative one.
Dec
16
comment Real life examples of commutative but non-associative operations
I was interpreting as $\circ$ takes two people and outputs their child. Then $(a\circ b)\circ c$ is the child made by (the child of $a$ and $b$) and $c$. And $a\circ(b\circ c)$ is the child of $a$ and (the child of $b$ and $c$). Taking $a$, $b$, and $c$ as the top nodes, you may draw the family tree, and see that in terms of DNA, the operation is not associative. I suppose it is not incest actually though, but does suggest a wide age gap between couples.
Dec
15
comment Prove that no $n,m, 0<n<m$ exist such that $m^2 +mn+n^2$ is a square number
I deleted a (wrong) proof I had to the contrary. I will put it here instead: View the expression as a polynomial in $\mathbb{Z}[m]$. If it is irreducible over $\mathbb{Q}$, clearly it cannot be factored as a perfect square. If we reduce modulo $2$, we find that $n$ is even, otherwise the polynomial is irreducible. Similarly, we can find the $m$ is even. We can then factor a $4$ out of the whole expression, and we must still get a perfect square (since $4$ is a perfect square itself). Let $m' = m/2$ and $n' = n/2$. We get that $m'^2 + m'n' + n'^2$ is also a perfect square. Repeat to infinity.