document.write( "Question 91198: Prove that cube root of 7 is an IRRATIONAL number. \n" ); document.write( "

Algebra.Com's Answer #623331 by ikleyn(52798) $\"\"$ $\"About$

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document.write( "Let us suppose that  is a rational number. Then we can write  = , where  and  are whole numbers. \r\n" );
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document.write( "We can assume that the numbers  and  have no common divisor. Otherwise, we can cancel this common divisor \r\n" );
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document.write( "in the numerator and denominator. \r\n" );
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document.write( "Raise both sides of the equality  =  in the degree 3. You will get  = . Hence,  = .\r\n" );
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document.write( "The right side of the last equality is divisible by . Hence, its left side  is divisible by  also. \r\n" );
document.write( "Since  is divisible by , the whole number  itself is divisible by . \r\n" );
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document.write( "So, we can write  = , where  is another whole number. \r\n" );
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document.write( "Now, substitute  =  into  = . It gives  = . Cancel both sides by . You will get  = . \r\n" );
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document.write( "The left side of the last equality is divisible by . Hence, its right side  is divisible by  also. \r\n" );
document.write( "Since  is divisible by , the whole number  itself is divisible by .\r\n" );
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document.write( "We just got a contradiction. We assumed that  = , where  and  are whole numbers with no common divisor, and the chain \r\n" );
document.write( "of arguments led us to the conclusion that both the integers  and  have the common divisor . \r\n" );
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document.write( "This contradiction proves that the original assumption was wrong, regarding  as a rational number. Hence,  is irrational. \r\n" );
document.write( "The proof is completed. \r\n" );
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document.write( "In the proof we used many times this property of the number 7: if 7 divides the product of two integers m and n, then 7 divides \r\n" );
document.write( "at least one of the integers. It is the common property of any prime number in the ring of integer numbers.\r\n" );
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