MHB Rational Inequalities: Solve & Understand | Math

Achi_kun
Messages
5
Reaction score
0
E5C700FB-2211-45B3-8368-F2318DAF4F6B.jpeg
A1B4D6DC-C9F9-45EB-A985-E49539667A1B.jpeg
 
Mathematics news on Phys.org
Uhh ... fill in the blanks? Have you worked on any of these?
 
skeeter said:
Uhh ... fill in the blanks? Have you worked on any of these?
Idk how the solution works
 
for $\dfrac{1}{x} > 2$

step 1. $\dfrac{1}{x} - \dfrac{2}{1} > 0$

combine the two fractions by using a common denominator
 
Personally, I wouldn't do it that say.
From $\frac{1}{x}> 2%$, multiply on both sides by x.
But you have to be careful with that! Unlike with an equation, multiplying on both sides by negative number reverses the ">" sign. So do two cases:

1) If x> 0 then $1>2x$. Divide on both sides by the positive number 2: $\frac{1}{2}> x$..
Since we are requiring that x be positve, we have $0< x< \frac{1}{2}$

2) If x< 0 then $1< 2x$. Divide on both sides by the positive number 2: $\frac{1}{2}< x$, But since we are requiring that x be negative, that is not possible.

The solution is $0<x \frac{1}{2}$.

It is also true that, for continuous functions, g and f, to change from f(x)<g(x) to f(x)> g(x), we have to go through f(x)= g(x) or a poinr where either f or g is undefined.

So start by solving the equation $\frac{1}{x}= 2$. That is the same as $1= 2x$, or $x=\frac{1}{2}$. I is also true that $\frac{1}{x}$ is undefined for x= 0. That divides the real numbers into three intervals, x< 0, 0< x< 1/2, and x> 1/2. We need only check one value of x in each interval. For x< 0 take x=-1. Then 1/x= -1 which is NOT larger than 2 so no x less than 0 satisfies 1/x> 2. For 0< x< 1/2 we can take x=1/4. Then 1/x= 4 which is greater than 2. Every number betwen 0 and 1/2 satisfies the inequalty. Finally take x= 1. Then 1/x=1 which is not larger than 2. No x larger than 1/2 satisfies the inequality.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Back
Top