document.write( "Question 1023501: List all elements of the set A ∩ B, where
\n" ); document.write( "A = {n ∈ N | ∃k ∈ N such that n = 2^k + 2},
\n" ); document.write( "B = {n ∈ N | ∃k ∈ N such that n = 2^k − 2}. \n" ); document.write( "

Algebra.Com's Answer #639046 by robertb(5830) $\"\"$ $\"About$

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The first few elements of A are
\n" ); document.write( " $\"2%5E0%2B2+=+3\"$ , $\"2%5E1%2B2+=+4\"$ , $\"2%5E2%2B2+=+6\"$ , $\"2%5E3%2B2+=+10\"$ , $\"2%5E4%2B2+=+18\"$ , $\"2%5E5%2B2+=+34\"$ ,...
\n" ); document.write( "The first few elements of B are
\n" ); document.write( " $\"2%5E1-2=+0\"$ , $\"2%5E2-2+=+2\"$ , $\"2%5E3-2+=+6\"$ , $\"2%5E4-2+=+14\"$ , $\"2%5E5-2+=+30\"$ ,...\r
\n" ); document.write( "\n" ); document.write( "It is quite clear that 6 is a common element, so 6 is in A ∩ B.\r
\n" ); document.write( "\n" ); document.write( "Now we show that for k and $\"l%3E=4\"$ , there are no other common terms.\r
\n" ); document.write( "\n" ); document.write( "Suppose there are, or suppose there are natural numbers a, $\"b%3E=4\"$ such that \r
\n" ); document.write( "\n" ); document.write( " $\"2%5Ea+%2B+2+=+2%5Eb+-+2\"$
\n" ); document.write( "==> $\"2%5Ea+%2B+4+=+2%5Eb\"$ ,
\n" ); document.write( "==> $\"2%5E%28a-2%29%2B1++=+2%5E%28b-2%29\"$ , contradiction, because $\"2%5E%28a-2%29\"$ and $\"2%5E%28b-2%29\"$ would both be even since a, $\"b%3E=4\"$ .\r
\n" ); document.write( "\n" ); document.write( "Therefore A ∩ B = {6}.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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