prove that sqrt(5) is irrational :)

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- anonymous

prove that sqrt(5) is irrational :)

- jamiebookeater

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- anonymous

it's not a perfect square, so it's irrational

- anonymous

how to prove this?

- anonymous

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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## More answers

- anonymous

i know this already. but if you were given this question to work out. how would you do it?

- asnaseer

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.

- anonymous

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.

- asnaseer

all steps are correct so far :)

- asnaseer

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 \(a^2\)

- asnaseer

yo found:\[a^2=5b^2\]therefore \(a^2\) must be a multiple of ???

- anonymous

multiple of 5

- asnaseer

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?

- anonymous

yes...

- asnaseer

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.

- anonymous

i get \[25m ^{2} = 5b ^{2}\]

- asnaseer

yup - now divide both sides by 5 and see what you can say about \(b^2\)...

- anonymous

\[5m ^{2} = b ^{2}\]

- asnaseer

... so \(b^2\) must be a multiple of ???

- anonymous

or 5, just like \[a ^{2}\] was

- anonymous

of**

- asnaseer

yes - and, following the same reasoning as we did for \(a^2\), we conclude that b is also a multiple of 5.

- asnaseer

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.

- asnaseer

I hope that makes sense?

- anonymous

makes sense if we can say that a rational number is one which when expressed as a fraction has no coprime (except 1)

- asnaseer

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."

- anonymous

makes sense. thanks! goodnight!

- asnaseer

yw :)
and goodnight to you too.

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