The sum of two numbers is 53 and the difference is 11. What are the numbers?

1 answer:

4 0

Let assume that the the two numbers are X and Y and that X is the large number.
According to the question,
X + Y = 53 .................. Equation 1
then, X = 11 + Y........... Equation 2
Substitute equation 2 into equation 1.
[11 + Y] + Y = 53
11 + 2Y = 53
2Y = 53 - 11 = 42
2Y = 42
Y = 42 / 2 = 21,
Therefore, Y = 21 [Small number].
According to equation 1,
X +Y = 53
X + 21 = 53
X = 53 - 21 = 32.
Therefore, X = 32 [Big number].

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

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By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .