document.write( "Question 1130704: Using proof by contradiction, prove that 7√2 is irrational. \n" ); document.write( "

Algebra.Com's Answer #747326 by MathLover1(20849) $\"\"$ $\"About$

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\n" ); document.write( " suppose that $\"7sqrt%282%29\"$ were rational:\r
\n" ); document.write( "\n" ); document.write( "Then we can write $\"7sqrt%282%29\"$ as a fraction $\"+a%2Fb+\"$ where $\"+a\"$ and $\"b\"$ are integers with no common factors.\r
\n" ); document.write( "\n" ); document.write( "Since $\"7sqrt%282%29+=+a%2Fb\"$ , square both sides=> $\"7%5E2%2A2+=+a%5E2%2Fb%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "So, $\"7%5E2%2A2b%5E2+=+a%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "By the definition of even, this means $\"a%5E2\"$ is even. But then $\"a\"$ must be even if $\"a%5E2\"$ is even . \r
\n" ); document.write( "\n" ); document.write( "So $\"a+=+2n\"$ for some integer $\"n\"$ .
\n" ); document.write( "If $\"+a+=+2n\"$ and $\"7%5E2%2A2b%5E2+=+a%5E2\"$ , then \r
\n" ); document.write( "\n" ); document.write( " $\"7%5E2%2A2b%5E2+=+4n%5E2\"$ . \r
\n" ); document.write( "\n" ); document.write( "So $\"7%5E2%2Ab%5E2+=+2n%5E2\"$ . This means that $\"b%5E2\"$ is even, so $\"b\"$ must be even.\r
\n" ); document.write( "\n" ); document.write( "We now have a contradiction. $\"a\"$ and $\"b\"$ were chosen not to have any common factors. \r
\n" ); document.write( "\n" ); document.write( "But they are both even, i.e. they are both divisible by $\"2\"$ . \r
\n" ); document.write( "\n" ); document.write( "Because assuming that $\"7sqrt%282%29\"$ was rational led to a $\"contradiction\"$ , it must be the case that $\"7sqrt%282%29\"$ is irrational. \r
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