Waste less time on Facebook — follow Brilliant.
Common Misconceptions (Algebra)
\[\large \sqrt{-2}\times\sqrt{-3}=\ ?\]
In this problem, the square root is a function from the complex numbers to the complex numbers.
by
Omkar Kulkarni
If \(x\) and \(y\) are non-zero numbers such that \(x>y\), which of the following is always true?
(A) \(\dfrac{1}{x}<\dfrac{1}{y}\)
(B) \(\dfrac{x}{y}>1\)
(C) \(|x|>|y|\)
(D) \(\dfrac{1}{xy^2}>\dfrac{1}{x^2y}\)
(E) \(\dfrac{x}{y}>\dfrac{y}{x}\)
by
Khang Nguyen Thanh
Calvin has a collection of special weighted dice that all share special properties:
- They're all 4 sided dice
- Any one die has distinct positive integers on each of its faces
- No pair of dice have all their numbers exactly the same
- The probability of rolling number \(x\) on any one of the dice is \(\frac {1}{x}\)
Let \(a\) be the maximum number of dice in the collection and let \(S\) be the sum of all the faces of all the dice in the maximum collection size.
Find \(a+S\).
by
Sharky Kesa
I claim that \(1 < -1\) using the proof below but I was just told that I might have just committed a small mistake.
A. \(- ( i^4 - i^3 - i - 1)< 1 -i^2\)
B. \(i^3 - i^4 + i + 1 < 1-i^2\)
C. \((i^3+i)(1-i) < (1-i)(1+i) \)
D. \( i^3 + i < 1+i\)
E. \(i^2(1+i) < 1+i \)
F. \( i^2 < 1 \)
G. \( i^2 < i^4\)
H. \( 1 < i^2 \)
I. \( 1 < -1 \)
Where did I go wrong? At which step(s) did I commit a fallacy?
Note: \(i = \sqrt {-1} \).
by
Efren Medallo
What is the wrong step in the following proof that \(1 = -1\)?
Let \(w\) be a complex number such that \((w + 1)^3 = (w - 1)^3\).
Solving this equation gives \(w = \pm \frac{i \sqrt{3}}{3}\).
Since \((w + 1)^3 = (w - 1)^3\) for our previously mentioned values of \(w\), cube rooting both sides gives \(w + 1 = w - 1\).
Subtracting \(w\) from both sides gives \(1 = -1\).
by
Steven Yuan