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UkoKoshka [18]
3 years ago
5

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

b2≤c2+d2. Show that ∼ ∼ is reflexive and transitive but not symmetric.
Mathematics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $ \forall (a, b) \in \mathbb{R}^2 $, $ (a, b) R(a, b) $.

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$ (a, b) \sim (a, b) $ since $ a^2 + b^2 = a^2 + b^2 $

This is true for any pair of numbers in $ \mathbb{R}^2 $. So, $ \sim $ is reflexive.

Symmetry:

$ \sim $ is symmetry iff whenever $ (a, b) \sim (c, d) $ then $  (c, d) \sim (a, b) $.

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $

$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $

Hence, $ (a, b) \sim (c, d) $ since $ a^2 + b^2 \leq c^2 + d^2 $

Note that $ c^2 + d^2 \nleq a^2 + b^2 $

Hence, the given relation is not symmetric.

Transitive:

$ \sim $ is transitive iff whenever $ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $ then $ (a, b) \sim (e, f) $

To prove transitivity let us assume $ (a, b) \sim (c, d) $ and $ (c, d) \sim (e, f) $.

We have to show $ (a, b) \sim (e, f) $

Since $ (a, b) \sim (c, d) $ we have: $ a^2 + b^2 \leq c^2 + d^2 $

Since $ (c, d) \sim (e, f) $ we have: $ c^2 + d^2 \leq e^2 + f^2 $

Combining both the inequalities we get:

$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $

Therefore, we get:  $ a^2 + b^2 \leq e^2 + f^2 $

Therefore, $ \sim $ is transitive.

Hence, proved.

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3 years ago
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KonstantinChe [14]

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18

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x² = 16² + 30² - 2(16)(30)cos 30°

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Please mark my answer as brainliest

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3 years ago
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