Mathematics Asked by M. Choy on January 26, 2021
Let $x>1$. I’m trying to prove that the function $y=frac{1}{x+1}$ is positive. I’ve managed to prove that:
$$frac{1}{x+1}=frac{1}{x} + 1$$
$$frac{1}{x} > 0$$
$$frac{1}{x} + 1 > 1$$
However, how do I prove that $frac{1}{x}>0$ is positive?
You have $y = frac{1}{x + 1}$.
The numerator, $1$, is positive. The denominator, $x + 1$, is positive because it's said that $x > 1$, so $x + 1 > 1$.
Then $y$ is the quotient of two positive numbers, so $y$ must be positive.
Correct answer by Sigma on January 26, 2021
Answer:
*. $X>1Rightarrow X+1>2>0$
**. Also $1>0$
Depending (*, **), we find$frac{1}{X+1}>0 $
Answered by Anas anas on January 26, 2021
For any $aneq 0$ then $atimes frac 1a=1>0$ thus $a$ and $frac 1a$ have the same sign.
Thus $frac 1{x+1}>0iff x+1>0iff x>-1$ and you are done.
Answered by zwim on January 26, 2021
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