Proving a+br is Irrational Using Proof by Contradiction

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SUMMARY

The discussion centers on proving that if \( a \) and \( b \) are rational numbers (with \( b \neq 0 \)) and \( r \) is an irrational number, then \( a + br \) is irrational. The proof by contradiction begins with the assumption that \( a + br \) is rational, leading to a contradiction by demonstrating that \( r \) must also be rational. Key concepts include the closure properties of rational numbers under addition and multiplication, and the existence of a rational multiplicative inverse for non-zero rational numbers.

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  • Understanding of rational and irrational numbers
  • Familiarity with proof by contradiction
  • Knowledge of closure properties of rational numbers
  • Basic algebraic manipulation of equations
NEXT STEPS
  • Study the properties of rational and irrational numbers in depth
  • Learn more about proof techniques, specifically proof by contradiction
  • Explore the concept of closure in mathematical sets
  • Investigate the implications of multiplicative inverses in rational numbers
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Mathematicians, students studying number theory, educators teaching proofs, and anyone interested in the properties of rational and irrational numbers.

mr_coffee
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Hello everyone! THe directions are the following:
Carefully formulate the negations of each of the statements. Then prove each statement by contradiction.

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

I made it into a universal statement:
\forall real numbers a and b and r, if a nd b are rational such that b != 0, and r is irrational, then a + br is rational.

I then took the negation:
\exists rational numbers a and b, b != 0 and irrational number r such that a + br is rational.

Proof:
Suppose not. That is, suppose there are rational numbers a and b, with b != 0 and r is an irrational number such that a + br is rational. By definition of rational a + br = e/f and a = m/n and b = x/y for some integers e, f, m, n, x, and y with n,y,x, and f != 0. By substitution

a + br = e/f
a = m/n;
r = irrational
b = x/y

m/n + (x/y)*r = e/f

r = y/x(e/f – m/n)

r = (yen-fmy)/fnx

now yen-fmy and fnx are both integers, and fnx != 0. Hence r is a quotient of the integers yen-fmy and fnx with fnx != 0. Thus, by definition of rational, r is rational, which contradicts the supposition that a+br is irrational.


I know there are a lot of variables, but i wasn't sure how else to show that r is infact a quotient of integers.

Any comments on what i did if its correct or flawed? Thank you!
 
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What you've done is perfectly correct but you are right- it's hard to read with all those letters.

You might just use the fact that the set of rational numbers is closed under addition and multiplication- and every non-zero rational number has a rational multiplicative inverse.

If a+ br= c then r= (c-a)(1/b). If a, b, c, b not 0, are all rational then c-a is rational (because the rational numbers are closed under addition), 1/b is rational (b is not 0 so has a rational multiplicative inverse), (c-a)(1/b) is rational (the rational numbers are closed under multiplication).
 
That does look a lot nicer, thanks for the help! :biggrin:
 

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