MTH645 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || FUNCTIONAL ANALYSIS || BY VuTech

MTH645 ASSIGNMENT NO. 1 FALL 2022

MTH645 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || FUNCTIONAL ANALYSIS || BY VuTech

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Question

Let  X = {2, 4, 6, … ,20} be the Universe of discourse, 

`\A^\~ 1 = {\left( 2,0.5 \right),\left( 4,0.9 \right),\left( 6,0.2 \right),\left( 8,0.4 \right),\left( 10,0.5 \right),\left( 12,0.2 \right),\left( 14,0.4} \right),\left( {16,0.0} \right),\left( 18,0.5 \right),\left( 20,0.8 \right)}`

And 

`\B^{ \~ 1} = \left\{ \left( 2,0.5 \right),\left( 4,0.5 \right),\left( 6,0.5 \right),\left( 8,0.2 \right),\left( 10,0.3} \right),\left(12,0.3 \right),\left( 14,0.5 \right),\left( 16,0.8 \right),\left( 18,0.2\right),\left(20,0.4 \right)}`

Be two fuzzy sets on X. Determine `\A^{ \~ 1} \cup B^{ \~ 1}`

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Solutions:

Here

X = {2, 4, 6, … ,20} be the Universe of discourse.

First, we find `\A^{ \~1}` and `\B^{ \~1}`

We know that

`\A^{ \1} = 1 - \mu A^{ \ 1}`

So,

`\A^{ \~1} = {\left( 2,0.5 \right),\left( 4,0.1 \right),\left( 6,0. \right),\left( 8,0.6 \right),\left( 10,0.5 \right),\left( 12,0.8 \right),\left( 14,0.6 \right),\left( 16,1.0 \right),\left( 18,0.5 \right),\left( 20,0.2 \right)}`

And 

`\B^{ \~1} = { \left( 2,0.5 \right),\left( 4,0.5 \right),\left( 6,0.5 \right),\left( 8,0.8 \right),\left( 10,0.7 \right),\left( 12,0.7 \right),\left( 14,0.5 \right),\left( 16,0.2 \right),\left( 18,0.8 \right),\left( 20,0.6 \right)}`

Now

`\A^{ \~ 1} \cup B^{ \~ 1`} =  `{\left( 2,0.5 \right),\left( 4,0.5 \right),\left( 6,0.8 \right),\left( 8,0.8 \right),\left( 10,0.7 \right),\left( 12,0.8 \right),\left( 14,0.6 \right),\left( 16,1.0 \right),\left( 18,0.8 \right),\left( 20,0.6 \right)}`

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SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

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