document.write( "Question 485329: prove that, square root of 2 is not a rational number. \n" ); document.write( "

Algebra.Com's Answer #331960 by MathLover1(20850) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "The proof that $\"sqrt%282%29\"$ is $\"irrational\"$ :\r
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\n" ); document.write( "\n" ); document.write( " Let's $\"suppose\"$ $\"sqrt%282%29\"$ were a $\"rational\"$ number. Then we can write it $\"sqrt%282%29=a%2Fb+\"$ where $\"a\"$ , $\"b\"$ are $\"whole\"$ numbers, $\"b\"$ $\"+not\"$ $\"+zero\"$ . \r
\n" ); document.write( "\n" ); document.write( "We additionally make it so that this $\"a%2Fb\"$ is simplified to the lowest terms, since that can obviously be done with any fraction.\r
\n" ); document.write( "\n" ); document.write( "It follows that $\"2+=+a%5E2%2Fb%5E2\"$ , or $\"+a%5E2+=+2+%2A+b%5E2\"$ . So the square of $\"a\"$ is an $\"even\"$ $\"number\"$ since it is $\"two\"$ $\"times\"$ $\"+something\"$ . \r
\n" ); document.write( "\n" ); document.write( "From this we can know that $\"a\"$ itself is also an $\"even\"$ number. Why? Because it can't be $\"odd\"$ ; if $\"a\"$ itself was $\"odd\"$ , then $\"+a+%2A+a\"$ would be $\"odd\"$ $\"+too\"$ . Odd number times odd number is always odd. \r
\n" ); document.write( "\n" ); document.write( "-if $\"a\"$ itself is an $\"even\"$ number, then $\"a\"$ is $\"2\"$ $\"+times\"$ some other whole number, or $\"a+=+2k\"$ where $\"k\"$ is this other number. We don't need to know exactly what $\"k\"$ is; it won't matter. Soon is coming the $\"contradiction\"$ :\r
\n" ); document.write( "\n" ); document.write( "If we substitute $\"a+=+2k\"$ into the original equation $\"2+=+a%5E2%2Fb%5E2\"$ , this is what we get:\r
\n" ); document.write( "\n" ); document.write( " $\"2=+%282k%29%5E2%2Fb%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"2=+%094k%5E2%2Fb%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"2%2Ab%5E2=+4k%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"b%5E2%09=+2k%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "This means $\"b%5E2\"$ is $\"even\"$ , from which follows again that $\"b\"$ itself is an $\"even\"$ $\"+number\"$ !\r
\n" ); document.write( "\n" ); document.write( "WHY is that a $\"contradiction\"$ ? Because we started the whole process saying that $\"a%2Fb\"$ is simplified to the lowest terms, and now it turns out that $\"a\"$ and $\"b\"$ would $\"both\"$ be $\"even\"$ . So $\"sqrt%282%29\"$ $\"CANNOT\"$ be $\"rational\"$ .\r
\n" ); document.write( "\n" ); document.write( "conclusion: $\"sqrt%282%29\"$ $\"HAS\"$ to be $\"irrational\"$ . \n" ); document.write( "

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