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lesantik [10]
3 years ago
6

Let A, B R be non-empty sets. Suppose that A B. Suppose that B has a least upper bound. Prove that A has a least upper bound, an

d lub A lessthanorequalto lub B.
Mathematics
1 answer:
Umnica [9.8K]3 years ago
8 0

Answer with explanation:

Given → A, B ⊆R, are non-empty sets.

Also, A ⊆ B.

It is also given that , B has least upper bound.

To Prove:→A has a least upper bound,

and → lub A ≤ lub B.

Proof:

    A and B are non empty sets.

Also, A ⊆ B

it means there are some elements in B which is not in A and there are some elements in set B which may be greater or smaller than or equal to set A.

The meaning of least upper bound is the the smallest member of the set from which all members are greater.

As, A⊆B

So, there are two possibilities ,either

→ Least upper bound set A = least upper bound set B------(1)

→ Least upper bound set A < least upper bound set B-------(2)

Combining (1) and (2)

→→→l u b(A) ≤ l u b (B)

   

 

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The equation in slope-intercept form for the line that passes through the point  ( -1 , -2 )  and is perpendicular to the line − 4 x − 3 y  =  − 5 is y = \frac{3}{4}x - \frac{5}{4}

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Given that the line that passes through the point  ( -1 , -2 )  and is perpendicular to the line − 4 x − 3 y  =  − 5

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\frac{-4}{3} \times \text{ slope of line perpendicular to it}= -1\\\\\text{ slope of line perpendicular to it} = \frac{3}{4}

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Substitute (x, y) = (-1, -2) and m = 3/4 in eqn 1

-2 = \frac{3}{4}(-1) + c\\\\-2 = \frac{-3}{4} + c\\\\c = - 2 + \frac{3}{4}\\\\c = \frac{-5}{4}

\text{ substitute } c = \frac{-5}{4} \text{ and } m = \frac{3}{4} \text{ in eqn 1}

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