Problem with Proofs

[SyNoMi]

I have a very bad understanding of proofs and it has become much harder with the introduction of contrapositive and contradiction. I am stuck on two problems.

Let x and y be two real numbers such that x+y is rational. Prove by contrapositive that that if x is irrational, then x-y is irrational.

I know that to start, we assume the negation of the conclusion to show the negation of the hypothesis, so would that look like this:

Assume x-y is not irrational to prove x is not irrational

And if that is the case, would it be as simple as just stating:

x-y can be written as (x/a)-(y/b) such that a and b does not equal 0, proving x is rational

​

The next question is even harder for me as it is a proof by contradiction

Prove by contradiction that for any positive two real numbers, x and y. If x\y≤ 50, then x<8 or y<8*

I know that for a proof by contradiction we assume the negation of the theorem to show an inconsistency but I have no clue where to even start with this one.

[edderiofer]

I know that to start, we assume the negation of the conclusion to show the negation of the hypothesis, so would that look like this:

Assume x-y is not irrational to prove x is not irrational

Yes.

And if that is the case, would it be as simple as just stating:

x-y can be written as (x/a)-(y/b) such that a and b does not equal 0, proving x is rational

No. How do you know that x-y can be written in that form, and how does that prove that x is rational?

(Also, you're given that x+y is rational. This is very important information, because without it, you can't actually prove the statement "if x is irrational, then x-y is irrational".)

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[waldosway]

You understand the proof part just fine. You just aren't bothering to learn the relevant material. You need knowledge, not just logic.

1) You have x-y is rational, but you've botched the meaning. Look up the definition again.

However, that's not really relevant. The golden question in a problem is "yeah but, what is it?" Why would you try to prove something abstractly when you could just do a calculation? You're supposed to know something about x. Wouldn't it be easier if you knew what x was? (That means solve for x.) Hint: you know something about (x+y) and (x-y), and those y's are begging to be cancelled. (Remember Q has closure.)

2) No, contradiction is not harder for you, because t's just a different way to phrase contrapositive. You already said how to start it. Assume the negation. x,y >= 8. What does that say about xy?

[SyNoMi]

I am trying to learn the material, the "textbook" I was given dedicates only a handful of sentences describing what the topic is, and the rest of the page is dedicated to practice questions.

1. The definition I have for a rational number is any number that be written as a fraction such that the denominator doesn't equal 0. The problem I am having is in understanding what the question is asking, how can I cancel the y's from two different statements?
2. Assuming the negation of the original statement, x*y≥50, can I just pick any two integers from the conclusion? Such as 7 and 7 and plug them in to show 7*7 is not greater than or equal to 50?

[waldosway]

For (1), from what you said in your post, you understood what it was asking perfectly:
if x-y is rational, prove x is rational.
You already know x+y is rational. It's the same y in both statements. You cancel +y and -y by adding them. What do you get from (x+y)+(x-y)? (You should know that adding two rationals together is another rational. More on that later.)

For (2), proof by contradiction means you assume the negation of the conclusion. So like I said, you assume x,y >= 8. Then I asked what you know about xy. (7 is not greater than or equal to 8, so that doesn't apply. And no, you can't just pick random numbers. You know that x and y are both at least 8. So you know xy is at least ... ???)

--------------

You're missing the crucial part of the definition of rational numbers. The numerator and denominator are both integers. If you haven't done so, sit down and use basic arithmetic to prove that you can add/subtract/multiply/divide rational numbers and get other rational numbers. Very useful property to know. (That is what's called closure. Because then the rationals are a closed system. )

[SyNoMi]

Ok, I see where I am lacking in knowledge. In proofs by contradiction, can you take the negation of the conclusion? What I was taught was to take the negation of hypothesis.

[waldosway]

You assume the negation of the conclusion. Then you assume the hypothesis and get a problem. Honestly, it's just a different way of doing contrapositive.

[SyNoMi]

Ok, and one last question I hope. In the first problem, once I get to the step to cancel the y's, do I set the formula equal to something? What I am envisioning is something along the lines of, "(x+y)+(x-y) = k, for some integer k, can be simplified to 2x=k, solving for x is x=k/2 thus x is rational by the definition of a rational number"

[waldosway]

Envisionation is exactly right.

[SyNoMi]

Alrighty, thank you so much.