Trying to prove by Contradiction, but need to make sure negation is correct

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The discussion focuses on proving a mathematical statement by contradiction, specifically the assertion that if \( a \) and \( b \) are rational numbers (with \( b \neq 0 \)) and \( r \) is an irrational number, then \( a + br \) is irrational. The user correctly translates the statement into a universal form and identifies the negation as the existence of rational numbers \( a \) and \( b \) (where \( b \neq 0 \)) and an irrational number \( r \) such that \( a + br \) is rational. The conversation emphasizes the importance of accurately formulating the negation in the proof process.

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mr_coffee
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Hello everyone. Before I start this proof i have to make sure my negation is right of the statement and also my translation of it into universal statements before I take the negation.

The directions says, prove the statements by method of contradition. The method of contradiction states:
1. Suppose the statement to be proved is false. that is, suppose that the negation of the statement is true. (be very careful in writing the negation!)
2. Show that this supposition leads logicaly to a contradiction.
3. Conclude that the statement to be proved is true.

Here is the problem:
If a and b are rational numbers, b != 0, and r is an irrational number, then a+br is irrational.

So here is me translating it into universal statements:
\forall real numbers a and b and r, if a and b are rational such that b !=0, and r is irrational, then a + br is irrational.

Here is me taking the negatoin of it:
\exists rational numbers a and b, b != 0 and irrational number r such that a + br is rational.

Does this look okay to you? The universal statement seemed odd, because i had so many ands in it but i wzsn't sure how else to write it.

Thanks!
 
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In your initial statement, why say that a,b are real and then restrict to Q? So, for all a,b in Q and r in R\Q: ...
 
Your right i guess i didn't need to say real numbers.
If i took out the real part, would i sitll end up with the same negation? It seems i would.

There exists rational numbers a,b, and b != 0 and irrational number r such that a+br is rational.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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