a) \(x + r\) is irrational
Proof:
Let \(x\) be irrational and \(r \in Q\).
By was of contradiction, prove \(x + r\) is rational.
Assume \(x + r = k\), where \(k ∈ Q\).
Since \(r\) and \(k\) are rational numbers, \(r = a/b\), \(k = c/d\) where \(∃ a, b, c, d ∈ Z, b, d ≠ 0\).
We can rewrite as \(x + a/b= c/d\) and simplify to \(x = c/d - a/b\) and further to \(x = (cb - ad)/db\).
This is a contradiction, \(x ≠ (cb - ad)/db\) since \(x\) is irrational and \((cb - ad)/db\) is a rational number (fraction).
So \(x + r\) must be irrational
b) If \(r \ne 0, xr \) is irrational
Proof.
Let \(x \) be irrational and \(r \in Q\).
By way of contradiction, \(xr\) is rational.
So we can assume \(xr = a/b\), where \(a, b \in \mathbb{Z}, b ≠ 0\).
Since \(r\) is rational, \(r = c/d\), where \(c, d \in Z, d \ne 0\).
We can rewrite as \(xc/d = a/b\) and simplify to \(x = a/b / cd \) and further to \(x = ad/cd\).
This is a contradiction, \(x \ne ad/cb\) since \(x\) is irrational and \(ad/cb\) is a rational number (fraction),
So \(xr\) must be irrational.
Looking good! Here are some tweaks to perfect your proof:
ReplyDeletePart (a):
(1) In the second line, "By was of contradiction" should be "By way of contradiction"
(2) "prove x+r rational" should be "suppose x+r rational"
(3) In the 4th line, "where \( \exists \)" is redundant - use one of the other
Part (b)
(4) In the second line, "By way of contradiction, xr is rational" should be "By way of contradiction, suppose xr is rational".
(5) Three lines from the end, \(x=a/b/cd\) is confusing to read. The formatting should be clarified. Note that if you want to type a fraction in LaTeX, you can use \frac{a}{b} which will come out as \( \frac{a}{b} \).
Nice work!
One extra comment - please use labels to tag your proof as a proof by contradiction and as involving rational numbers.
ReplyDelete