## tomiko Group Title prove that sqrt(5) is irrational :) one year ago one year ago

1. Sheng Group Title

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

2. tomiko Group Title

how to prove this?

3. Sheng Group Title

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

4. tomiko Group Title

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

5. asnaseer Group Title

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 Group Title

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 Group Title

all steps are correct so far :)

8. asnaseer Group Title

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 Group Title

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

10. tomiko Group Title

multiple of 5

11. asnaseer Group Title

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 Group Title

yes...

13. asnaseer Group Title

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 Group Title

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

15. asnaseer Group Title

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

16. tomiko Group Title

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

17. asnaseer Group Title

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

18. tomiko Group Title

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

19. tomiko Group Title

of**

20. asnaseer Group Title

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

21. asnaseer Group Title

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 Group Title

I hope that makes sense?

23. tomiko Group Title

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 Group Title

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 Group Title

makes sense. thanks! goodnight!

26. asnaseer Group Title

yw :) and goodnight to you too.