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ShengBest ResponseYou've already chosen the best response.0
it's not a perfect square, so it's irrational
 one year ago

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

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

asnaseerBest ResponseYou've already chosen the best response.1
One 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.
 one year ago

tomikoBest ResponseYou've already chosen the best response.0
if 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.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
all steps are correct so far :)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
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. now think about what the last step means  what does it say about \(a^2\)
 one year ago

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

asnaseerBest ResponseYou've already chosen the best response.1
correct. 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?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
therefore 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.
 one year ago

tomikoBest ResponseYou've already chosen the best response.0
i get \[25m ^{2} = 5b ^{2}\]
 one year ago

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

asnaseerBest ResponseYou've already chosen the best response.1
... so \(b^2\) must be a multiple of ???
 one year ago

tomikoBest ResponseYou've already chosen the best response.0
or 5, just like \[a ^{2}\] was
 one year ago

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

asnaseerBest ResponseYou've already chosen the best response.1
so 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.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I hope that makes sense?
 one year ago

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

asnaseerBest ResponseYou've already chosen the best response.1
yes  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."
 one year ago

tomikoBest ResponseYou've already chosen the best response.0
makes sense. thanks! goodnight!
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yw :) and goodnight to you too.
 one year ago
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