3 years ago

I need help on how to do this

Mathematics

1 answer:

Sholpan [36]3 years ago

5 0

C (48/6 is 8 and 8x2=16 16+74 (original temp) = 90 90-26=64)

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Salvador just purchased his first home and wants a nice television. He plans to keep the television for several years and has no

adelina 88 [10]

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a Hitachi 50 plasma tv at 495 per month for 6

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3 years ago

"assume that s is a non-empty set of real numbers which is bounded above and lambda is the least upper bound. Prove that for all

s2008m [1.1K]

By definition, if $\lambda$ is the least upper bound of the set $S$ , it means two thing:

$\forall x \in S,\ x \leq \lambda$
$\forall \varepsilon>0,\ \exists x \in S:\ x>\lambda-\varepsilon$

In other words, the least upper bound of a set is greater than or equal to every single element of the set, but it is "close enough" to the elements of the set, because you guaranteed to find elements in the set between $\lambda-\varepsilon$ and $\lambda$

For example, pick $S = [1,10)$ . Obvisouly, the least upper bound is $\lambda = 10$ . In fact, every number in $[1,10)$ is smaller than 10, but as soon as you take away something from 10, say 0.01, you get 9.99, and there are elements in $S$ greater than 9.99, say 9.9999.

So, the claim is basically proven by definition: if $\beta < \lambda$ , let $0 < \delta = \lambda - \beta$ . By definition, there exists $\alpha \in S:\ \alpha > \lambda - \delta$ .