- #1
mr_coffee
- 1,629
- 1
Hello everyone.
He told us he could of course change the parameters which he will of the proofs we have been working on so I'm testing out some cases but I want to make sure I'm doing it right.
Here is an example of a proof the boook had:
http://suprfile.com/src/1/3qaagxh/Untitled-1%20copy.jpg
Now where he has the statement: "If a and b are rational numbers..."
I'm changing that to:
"If a and b are integers..."
and now here is my proof, i think its correct but I have to make sure.
If a and b are integers, b != 0, and r is an irrational number, then a+ br is irrational.
Proof by Contradiction:
Suppose not. Suppose that a and b are integers, b != 0, and r is an irrational number such that a+br is rational. We must obtain a contradiction.
Since a, b are integers and a + br are rational, a+br = m/n for some integers m, n with n != 0.
Then
a + br = m/n
br = m/n - a
r = (m-an)/bn
where (m-an) and (bn) are integers since m, a, n, and b are integers, and bn is nonzero since b is nonzero. Therefore, r is rational, contradicting that r is irrational.
Thanks!
He told us he could of course change the parameters which he will of the proofs we have been working on so I'm testing out some cases but I want to make sure I'm doing it right.
Here is an example of a proof the boook had:
http://suprfile.com/src/1/3qaagxh/Untitled-1%20copy.jpg
Now where he has the statement: "If a and b are rational numbers..."
I'm changing that to:
"If a and b are integers..."
and now here is my proof, i think its correct but I have to make sure.
If a and b are integers, b != 0, and r is an irrational number, then a+ br is irrational.
Proof by Contradiction:
Suppose not. Suppose that a and b are integers, b != 0, and r is an irrational number such that a+br is rational. We must obtain a contradiction.
Since a, b are integers and a + br are rational, a+br = m/n for some integers m, n with n != 0.
Then
a + br = m/n
br = m/n - a
r = (m-an)/bn
where (m-an) and (bn) are integers since m, a, n, and b are integers, and bn is nonzero since b is nonzero. Therefore, r is rational, contradicting that r is irrational.
Thanks!
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