Prove that X ⊆ X U Y for all sets X and Y

  • Thread starter avdnowhere
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  • #1
avdnowhere
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hi guys...

this is my first thread in this forum..

i hope i'll learn math more easily with this forum...

mmm..

i've a some math homework that i must submit it this Thursday... :tongue2:

but, i don't know how to answer it...

the questions is...

1. Prove that X ⊆ X U Y for all sets X and Y...

2. A sequence r is defined as rn = 3.2ⁿ - 4.5ⁿ, n≥0.
Prove that the sequence satisfied rn = 7rn-1 - 10rn-2, n≥2.

anyone have the solution for these question???
 

Answers and Replies

  • #2
LCKurtz
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You need to show what you have tried on homework problems. Then people will give you hints or suggestions when they see where you are having difficulty. Welcome to the forums.
 
  • #3
avdnowhere
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You need to show what you have tried on homework problems. Then people will give you hints or suggestions when they see where you are having difficulty. Welcome to the forums.

i've a lecture note about this question...

but the note is different from the question...

so i don't know how to start it...
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,568
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hi guys...

this is my first thread in this forum..

i hope i'll learn math more easily with this forum...

mmm..

i've a some math homework that i must submit it this Thursday... :tongue2:

but, i don't know how to answer it...

the questions is...

1. Prove that X ⊆ X U Y for all sets X and Y...

2. A sequence r is defined as rn = 3.2ⁿ - 4.5ⁿ, n≥0.
Prove that the sequence satisfied rn = 7rn-1 - 10rn-2, n≥2.

anyone have the solution for these question???

i've a lecture note about this question...

but the note is different from the question...

so i don't know how to start it...

I will give you a couple hints. For the first one you must show every element in X is an element of X U Y. So start with an x in X and explain why it is in X U Y.

For the second one just calculate the right hand side and collect terms on powers of two and 5. It's straightforward algebra.
 

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