prove that sqrt(5) is irrational :)

Question

anonymous · Answer

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

anonymous · Answer

how to prove this?

anonymous · Answer

all square roots of numbers other than perfect squares yields an irrational response, because that's the definition of a perfect square

anonymous · Answer

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

asnaseer · Answer

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 · Answer

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 · Answer

all steps are correct so far :)

asnaseer · Answer

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 · Answer

yo found:$$a^2=5b^2$$therefore $a^2$ must be a multiple of ???

anonymous · Answer

multiple of 5

asnaseer · Answer

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 · Answer

yes...

asnaseer · Answer

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 · Answer

i get $$25m ^{2} = 5b ^{2}$$

asnaseer · Answer

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

anonymous · Answer

$$5m ^{2} = b ^{2}$$

asnaseer · Answer

... so $b^2$ must be a multiple of ???

anonymous · Answer

or 5, just like $$a ^{2}$$ was

anonymous · Answer

of**

asnaseer · Answer

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

asnaseer · Answer

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 · Answer

I hope that makes sense?

anonymous · Answer

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

asnaseer · Answer

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 · Answer

makes sense. thanks! goodnight!

asnaseer · Answer

yw :)

and goodnight to you too.