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tomiko
 3 years ago
prove that sqrt(5) is irrational :)
tomiko
 3 years ago
prove that sqrt(5) is irrational :)

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Sheng
 3 years ago
Best ResponseYou've already chosen the best response.0it's not a perfect square, so it's irrational

Sheng
 3 years ago
Best ResponseYou've already chosen the best response.0all square roots of numbers other than perfect squares yields an irrational response, because that's the definition of a perfect square

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0i know this already. but if you were given this question to work out. how would you do it?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1One way is to prove by contradiction. i.e. start off by assuming that \(\sqrt{5}\) IS rational and show that this leads to a contradiction  thus proving that it is irrational.

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0if i do this assumption. i can say that \[\sqrt{5} = \frac{ a }{ b }\] then \[5 = \frac{ a ^{2} }{ b ^{2} }\] \[5b ^{2} = a ^{2}\] and then i'm stuck.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1all steps are correct so far :)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1you also need to state at the start that a and b are coprime, i.e. they have no common factors so that a/b is a fraction in its simplest form. now think about what the last step means  what does it say about \(a^2\)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yo found:\[a^2=5b^2\]therefore \(a^2\) must be a multiple of ???

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1correct. now 5 is a prime number, so if \(a^2\) is a multiple of 5, then a must also be a multiple of 5  agreed?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1therefore we can express a as:\[a=5m\]where m is some other integer. therefore:\[a^2=(5m)^2=25m^2\]now substitute this back into your expression that relates a^2 to b^2 and see what that leads to.

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0i get \[25m ^{2} = 5b ^{2}\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yup  now divide both sides by 5 and see what you can say about \(b^2\)...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1... so \(b^2\) must be a multiple of ???

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0or 5, just like \[a ^{2}\] was

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yes  and, following the same reasoning as we did for \(a^2\), we conclude that b is also a multiple of 5.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1so we have concluded that both a and b are multiples of 5. which means they are NOT coprime, i.e. they have a common factor of 5. this contradicts our original assumptions, therefore \(\sqrt{5}\) must be irrational.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I hope that makes sense?

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0makes sense if we can say that a rational number is one which when expressed as a fraction has no coprime (except 1)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yes  that is how the proof is supposed to start off as  remember I stated it above as: "you also need to state at the start that a and b are coprime, i.e. they have no common factors so that a/b is a fraction in its simplest form."

tomiko
 3 years ago
Best ResponseYou've already chosen the best response.0makes sense. thanks! goodnight!

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yw :) and goodnight to you too.
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