rational number proof

[shrewd spade]

Is my proof ok?

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[shrewd spade]

(b) No. let a = 0 and b = sqrt(3) then ab = 0, which is rational.

[clear otter]

yes

[forest patrol]

Assuming a,b could be 0, then no

Otherwise for non-zero rational & irrational values, yes

[shrewd spade]

~~tex {C}
Yes. Let $a = \sqrt{\sqrt{2}}$, then $a ^ {2} = \sqrt{2}$, $a ^ {4} = 2$

[hollow stirrup]

anonymous190

[shrewd spade]

~tex {a}
If a is rational and b is irrational: suppose ${a+b} \in Q$, then $a+b = \frac{c}{d}$ where $c \in Z$ and $d \in Z$. Suppose $a = \frac{a_1}{a_2}$, where $a_1 \in Z$ and $a_2 \in Z$. Then $b = \frac{a_{2}c - da_1}{da_2} \in Q$.

[rancid hemlock]

anonymous190

[spiral plaza]

Indeed. Don't forget your conclusion, what does your work show, what's the final answer?

And do you have a counterexample of a+b being rational while a and b are both irrational?

[shrewd spade]

How to draw to conclusion?

[mental crow]

@shrewd spade

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[undone flume]

What do you mean?

You showed that b must be rational, when a+b and a are rational.

So.. the answer is obvious.

[mental crow]

@undone flume @spiral plaza @clear otter The user still needs help with this help request.

[undone flume]

?

[spiral plaza]

Well you just shown b in Q, but that wasn't the question, you assumed b is irrational. So what I means with not forgetting to write a conclusion is saying
"this is a contradiction, we assumed b was irrational so it can't be in Q, so a+b with a rational and b irrational cannot satisfy a+b in Q. So a sum of a irrational number and a rational number must be irrational."

Ofcourse I'm using a bit many words but you can't just end a proof with "b in Q" if you need to make a contradiction to actually finish your proof

[spiral plaza]

And the question (a) asks for the sum of a,b both irrational too.
Can you think of a counter-example such that a,b is irrational but a+b is rational, or can you proof this is impossible?

[shrewd spade]

Isn’t this proof by contrapositive?

[spiral plaza]

Look at your proof here. You start with "if b is irrational" and you finish with "b in Q" which means b is rational

So this is a proof by contradiction. You make a assumption, then proof that you get a contradiction, so the assumption must be wrong

[shrewd spade]

What’s the difference between proof by contradiction and contrapositive?

[spiral plaza]

I'd say intuitively it's pretty close. Contraposition is more strictly in logic, so really formulating a P⇒Q and then proving not Q ⇒ not P

I'd use the word contradiction when just writing out proofs in language, so when you say "we assume" and then disprove that assumption later, whereas with contrapositive I'd really start proving implications between logical statements

but quite possible some pedantic logicist can see this chat and angrily explain the exact difference between the 2 😃

[clever lodge]

SPiVaK!!

[shrewd spade]

+close

[mental crow]

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[mental crow]

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