*Proof* Sum of Rational and Irrational Numbers

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SUMMARY

The proof by contradiction establishes that if a and b are rational numbers (with b ≠ 0) and r is an irrational number, then the expression a + br is irrational. The negation of this statement leads to the conclusion that if a + br were rational, it could be expressed as a quotient of integers, contradicting the definition of irrational numbers. This proof relies on the properties of integers and the definitions of rational and irrational numbers, confirming the theorem's validity.

PREREQUISITES
  • Understanding of rational and irrational numbers
  • Familiarity with proof by contradiction
  • Basic algebraic manipulation of fractions
  • Knowledge of properties of integers
NEXT STEPS
  • Study the principles of proof by contradiction in mathematical logic
  • Explore the definitions and properties of rational and irrational numbers
  • Learn about algebraic manipulation involving fractions and integers
  • Review examples of similar proofs in number theory
USEFUL FOR

Students learning mathematical proofs, particularly in algebra and number theory, as well as educators seeking to clarify concepts related to rational and irrational numbers.

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Homework Statement


Prove by contradiction: If a and b are rational numbers and b != 0, and r is an irrational number, then a+br is irrational.

In addition, I am to use only properties of integers, the definitions of rational and irrational numbers, and algebra.

You guys should also know that I am new to proofs, so if I'm breaking convention in any blatant way, please let me know.

The Attempt at a Solution



First, I figure I need to find the negation of this statement:

Negation: There exists rational numbers a and b, b!=0, and irrational number r, such that a+br is rational.

I'm pretty sure this negation is correct, but I've been wrong before

Anyway, my proof would start like this:

Proof: To prove that for any rational numbers a,b,b!=0, and irrational number r, a+br is irrational, let's suppose not. Suppose there exists rational numbers a,b,b!=0 and irrational number r such that a+br is rational.
Let c,d,e,f be integers, by definition of rational, a+br can be rewritten as:

\frac{c}{d} + \frac{e}{f}*r is rational.

\frac{e}{f}*r = -\frac{c}{d}

r = -\frac{cf}{de}

\frac{cf}{de} can be written as a quotient of integers, so it is rational, therefore, our negation is false, therefore, our theorem is true?

I think I'm on the right track, but I know its messy and I might have made some logic mistakes...Also, my professor said that if we didn't have to explicitly say why b!=0, we probably weren't doing it right, and I never really did, so I have no idea...any help would be greatly appreciated.

edit-I guess if b was 0, that would make the quantity a+(br) 0, which leaves you with a, which is just a rational number
 
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In a proof by contradiction, you assume that the original hypothesis is true and that the conclusion is false.

Your original hypothesis is "a and b are rational numbers and b != 0, and r is an irrational number". Your original conclusion is "a + br is irrational".

The negative of your original conclusion is that a + br is not irrational, or in other words that a + br is rational.

Now start by assuming original hypothesis is true and that a + br is rational (the negated original conclusion). You should arrive at a contradiction, from which you can conclude that when the original hypothesis is true, the original conclusion is true, as well.
 

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