SafeX schreef:pompoen schreef:And anything multiplied by irrational number is irrational?
No, sqrt(2)*sqrt(2) is rational!
Ok, dank je. Ik vraag alleen omdat ik verschillende boeken/resources gebruik en ik heb gelezen dat "rational times irrational is irrational". Oh, and I guess I mismemorized "rational" as any number. Nevermind.
SafeX schreef:Why wouldn't it work with a rational integer? I mean, if you show that a numerator and a denominator have a common factor other than 1, (doesn't matter if its sqrt is rational or not), doesn't it create a contradiction already? Because the breuk could be further simplified? See what I mean?
Try to show that sqrt(4) is rational ...
Ik begrijp dat sqrt van 4 is een integer, daarom is het rational. Maar dat was niet mijn vraag.
]
SafeX schreef:
Yes, prove sqrt(2) is irrational ...
Umm, are you asking me to prove that sqrt of 2 is irrational? i guess you're imply that in working this one out I'll stumble upon an answer for my original question? Ok.(I'll do it in English since it's easier for me, hope it's ok.)
Let's assume the opposite. Let's assume sqrt(2) is rational. Than we can write it as a ratio:
sqrt(2)=a/b And we can also assume that a/b is irreducible (i.e. have no factors in common except for 1).
2=a^2/b^2
2b^2=a^2 --> that means a^2 (or axa) is even, meaning a is also even, because evenxeven=even.
If a is even it can be represented as 2 times some integer. a=2c.
2b^2=(2c)^2=4c^2
b^2=2c^2
c is an integer and any integer x2 is even. that means that b^2 is even, and then b is even (since, again evenxeven=even).
We assumed in the beginning that a and b are irreducible, but apparently they're not, if they're both even. They have a common factor 2. So that's a contradiction and sqrt(2) is not rational.
But I was wondering about the steps of working it out with a sqrt of a bigger number. Like, we factorize a number in the root and then use a factor that's irrational? But only if the factor is not cancelled/turned rational by another number, like in the case of sqrt(20)=(2x2x5), where sqrt(2x2) is rational so we ignore the twos and just use five? But if it were sqrt(2x5), could we use sqrt(2) then?
Also, if you had been snide, there was really no point in it. I'm obviously a noob and I'm just asking questions.