Proof: Show that 5 -√2 is an irrational number?

[doughishere]

Show that 5 -√2 is an irrational number?


So basically the text book has been telling me that the way to do these types of problems is to do these by proof-by-contradiction. I think the proof goes like this:

Assume that 5-√2 is a rational number. Thus, 5-√2 - 5 implies that -√2 is a rational number because the difference of two rational numbers is a rational number. We know this is not true, resulting in our original assumption to not be true also. In other words, 5 -√2 is an irrational number.


Is this right?

 

[ksdhart2]

Everything looks good to me. The rational numbers are, indeed, closed under subtraction, and we know that \(\displaystyle \sqrt{2}\) is not a rational number. It's a very quick and easy proof. Sometimes it's a sign you've messed up when you get a result that just seems to be too easy or too neat. But, this isn't one of those times.

The only possible tiny thing I could think of to improve it is to go all "first principles" and independently prove that \(\displaystyle \sqrt{2}\) is irrational, and that the rational numbers are closed under subtraction. But, depending on your level of math, this is probably wholly unnecessary.

 

[doughishere]

Everything looks good to me. The rational numbers are, indeed, closed under subtraction, and we know that \(\displaystyle \sqrt{2}\) is not a rational number. It's a very quick and easy proof. Sometimes it's a sign you've messed up when you get a result that just seems to be too easy or too neat. But, this isn't one of those times.

The only possible tiny thing I could think of to improve it is to go all "first principles" and independently prove that \(\displaystyle \sqrt{2}\) is irrational, and that the rational numbers are closed under subtraction. But, depending on your level of math, this is probably wholly unnecessary.

Thanks. I think the proof is that no rational number has a square root equal to 2....this is a question in the opening section of sheldon axlers text book.

The concept of closure that you talk about.....my understanding of it is this.....a set is closed if and only if the operation leads to a result that is in the set also. This works fine for addition subtraction and multiplication if the set is all rational number...why is division not....is division just not all the time or is it all the time....for instance 8 divided by 8 gives a rational number 1/1 or 1...but 3/0 is not a rational number...its undefined...is division and this closure.....conditional on the specific operation?

or am i just way off.

 

[tkhunny]

Closed mean Closed. One contrary example is sufficient to prove it is NOT Closed.

Natural Numbers are NOT Closed for Subtraction, for example. 3 - 7 = -4. -4 < 1 -- Nope.

Integers are NOT Closed for Division, for example. 6/24 = 1/4. 0 < 1/4 < 1 -- Nope.

 

[doughishere]

Closed mean Closed. One contrary example is sufficient to prove it is NOT Closed.

Natural Numbers are NOT Closed for Subtraction, for example. 3 - 7 = -4. -4 < 1 -- Nope.

Integers are NOT Closed for Division, for example. 6/24 = 1/4. 0 < 1/4 < 1 -- Nope.

I meant real numbers for my example.

 

[doughishere]

I meant real numbers for my example.

Back to the proof. I think the proof is wrong because the result is \(\displaystyle -\sqrt{2}\) and it should be \(\displaystyle \sqrt{2}\)...thus the math part should look like \(\displaystyle 5-(5-\sqrt{2})=\sqrt{2}\)

 

[doughishere]

Doug, distresses me to see you still sleeping in that church...

its just my icon. Im just trying to learn...sometimes I get so bummed with how little I know in life and how small amount of a time I have to try and learn it all....this comes easy for a lot of people. not me. but i keep trying and keep wanting to know more and i want to know if im correct or not so i can adjust my thinking.....anyways.

Ive cleaned it up a little bit.


Show that \(\displaystyle 5-\sqrt{2}\) is an irrational number.

Proof: Suppose \(\displaystyle 5-\sqrt{2}\) is a rational number. Because \(\displaystyle \sqrt{2}=5-(5-\sqrt{2})\) implies that \(\displaystyle \sqrt{2}\) is a rational number via the property that the difference of two rational numbers is also a rational number. We know \(\displaystyle \sqrt{2}\) is not rational and thus by counter example, \(\displaystyle 5-(5-\sqrt{2})\) is irrational.

 

[ksdhart2]

Strictly speaking, either formulation is fine. The proclamation that \(\displaystyle -\sqrt{2}\) is irrational suffices, because if any number is irrational, its negative must also be irrational. But it never hurts to be a bit more explicit and clear.

As for the closure idea, your concerns about division by zero are valid. It may seem as though no set containing 0 can ever be closed under division, because x/0 isn't a real number. However, we get around that by defining division in terms of multiplication by the inverse. In other words, we say that a/b = a * (1/b). So, for the real numbers to be closed under division, what we're really after is: Is it closed under multiplication by the subset of all real numbers of the form 1/b. And since 1/b is not a real number when b = 0 (in fact, it's completely undefined, so it doesn't exist in any context), we don't need to consider it. Hopefully that helps alleviate your confusion a little.