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prove that sqrt(5) is irrational :)

Mathematics
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it's not a perfect square, so it's irrational
how to prove this?
all square roots of numbers other than perfect squares yields an irrational response, because that's the definition of a perfect square

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Other answers:

i know this already. but if you were given this question to work out. how would you do it?
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.
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.
all steps are correct so far :)
you also need to state at the start that a and b are co-prime, 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 \(a^2\)
yo found:\[a^2=5b^2\]therefore \(a^2\) must be a multiple of ???
multiple of 5
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?
yes...
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.
i get \[25m ^{2} = 5b ^{2}\]
yup - now divide both sides by 5 and see what you can say about \(b^2\)...
\[5m ^{2} = b ^{2}\]
... so \(b^2\) must be a multiple of ???
or 5, just like \[a ^{2}\] was
of**
yes - and, following the same reasoning as we did for \(a^2\), we conclude that b is also a multiple of 5.
so we have concluded that both a and b are multiples of 5. which means they are NOT co-prime, i.e. they have a common factor of 5. this contradicts our original assumptions, therefore \(\sqrt{5}\) must be irrational.
I hope that makes sense?
makes sense if we can say that a rational number is one which when expressed as a fraction has no coprime (except 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 co-prime, i.e. they have no common factors so that a/b is a fraction in its simplest form."
makes sense. thanks! goodnight!
yw :) and goodnight to you too.

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