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

1. Sheng

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

2. tomiko

how to prove this?

3. Sheng

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

4. tomiko

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

5. 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.

6. tomiko

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.

7. asnaseer

all steps are correct so far :)

8. 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 $$a^2$$

9. asnaseer

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

10. tomiko

multiple of 5

11. 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?

12. tomiko

yes...

13. 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.

14. tomiko

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

15. asnaseer

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

16. tomiko

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

17. asnaseer

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

18. tomiko

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

19. tomiko

of**

20. asnaseer

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

21. 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.

22. asnaseer

I hope that makes sense?

23. tomiko

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

24. 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."

25. tomiko

makes sense. thanks! goodnight!

26. asnaseer

yw :) and goodnight to you too.