Show that if then or for all integers . I don't know how to start.
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Originally Posted by math2011 Show that if then or for all integers . I don't know how to start. suppose where then : , hence : If is an even number then , contradiction..... If is an odd number then are even numbers , hence : , contradiction.....
Last edited by princeps; Mar 9th 2012 at 11:35 PM.
Hello. You could do following: Let be such that , then we're goning to show that . If , and , so . On the other hand, . Hence , but , then therefore or Best regards.
Thank you! This is the part that I could not prove.
Originally Posted by Fernando Hello. You could do following: Let be such that , then we're goning to show that . If , and , so . On the other hand, . Hence , but , then therefore or Best regards. I don't fully understand the logic in your proof. Do you mean when you say ? Why does and imply ?
Hi, yeah ! it's my notation. OK. There is a theorem states that: Let and be integers differents of zero. Then iff holds following: and If and then . So, by the theorem We can assert if and then and I guess you could continue... Best regards.
Last edited by Fernando; Mar 11th 2012 at 07:11 AM. Reason: Grammar
Originally Posted by Fernando Hi, yeah ! it's my notation. OK. There is a theorem states that: Let and be integers differents of zero. Then iff holds following: and If and then . So, by the theorem We can assert if and then and I guess you could continue... Best regards. Do you mean the following theorem? Let and not both zero. Then a positive integer is the greatest common divisor of and if and only if the following equivalence holds: . This is different to what you stated.
Hi I don't know, what's the wrong? Do you think the first theorem is wrong?
Originally Posted by Fernando Hi I don't know, what's the wrong? Do you think the first theorem is wrong? I think the first theorem is correct after thinking through it. Sorry I had not seen it before and confused it with the one that I posted.
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