## tomiko 2 years ago how do i work out which number is greater? (sqrt12 - sqrt11), (sqrt13 - sqrt12)

1. asnaseer

@tomiko are you aware of this factorisation?$a^2-b^2=(a+b)(a-b)$

2. tomiko

3. asnaseer

look at the first term:$\sqrt{12}-\sqrt{11}$if you multiply this by $$\sqrt{12}+\sqrt{11}$$ what will you get?

4. tomiko

you get 1

5. asnaseer

correct. nor similarly multiply the second expression with $$\sqrt{13}+\sqrt{12}$$ - what do you get?

6. asnaseer

*now

7. tomiko

for that too i get 1. but one is definately greater. cause when i use the calculate i get 2 different values, making one of the pairs greater than the other.

8. asnaseer

good. now, out these two terms, which one is bigger:$1. \sqrt{12}+\sqrt{11}$$2.\sqrt{12}+\sqrt{13}$

9. tomiko

$\sqrt{12} + \sqrt{13}$ is bigger

10. asnaseer

good. so you now know that: a) x times (1) = 1 b) y times (2) = 1 and you know (2) is bigger than one. therefore what can you conclude about x and y?

11. asnaseer

I meant to say: you know (2) is bigger than (1)

12. tomiko

yes. 2 is bigger than one so that makes $\sqrt{13} - \sqrt{12}$ bigger?? :D

13. asnaseer

no - in terms of x and y that I listed, x is being multiplied by a smaller number than y is being multiplied by, but both end up as 1. therefore is x or y bigger?

14. tomiko

in that case x is bigger

15. asnaseer

imagine you had:$x*4=1$$y*8=1$which one would you say is bigger, x or y?

16. tomiko

x

17. asnaseer

yes, x is bigger, therefore you know which one is bigger in your original question

18. tomiko

yes i know $\sqrt{13} - \sqrt{12}$ is bigger.

19. asnaseer

20. asnaseer

remember that $$\sqrt{13} - \sqrt{12}$$ was multiplied by the bigger number $$\sqrt{13} + \sqrt{12}$$ to get 1.

21. tomiko

yes!! i get it now. had to look at my solutions again using the different of 2 squares.

22. asnaseer

we showed that:$(\sqrt{12}-\sqrt{11})(\sqrt{12}+\sqrt{11})=1$$(\sqrt{13}-\sqrt{12})(\sqrt{13}+\sqrt{12})=1$and$\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}$

23. tomiko

that makes $\sqrt{12} - \sqrt{11}$ greater!

24. asnaseer

perfect! well done! :)

25. tomiko

i undertand now. thank you very much! but you have any idea about this approach? Purely algebra ? Place ? i/o =, > or <. sqrt(12)-sqrt(11) ? sqrt(13)-sqrt(12) 2sqrt(12) ? sqrt(11)+sqrt(13) 48 ? ... Can you continue ?

26. asnaseer

ok what this approach is doing is bringing together the two square roots of 12 and and then squaring both sides. so:$(2\sqrt{12})^2=4*12=48$now you just need to square the right hand side:$(\sqrt{11}+\sqrt{13})^2=?$

27. asnaseer

do you understand?

28. tomiko

yeap...! i'm trying to do it on paper here :D

29. asnaseer

ok :)

30. tomiko

and this shows $\sqrt{12} - \sqrt{11}$ is bigger. you can see my working attached as an image

31. asnaseer

you expansion of $$(\sqrt{11}+\sqrt{13})^2$$ is not correct. remember that:$(a+b)^2=a^2+2ab+b^2$

32. tomiko

so if i correct it. i get $48 ? 24 + 2\times \sqrt{13}\times \sqrt{11}$

33. asnaseer

that is correct so far. now take the 24 to the left hand side and then divide both sides by 2

34. asnaseer

also remember that:$\sqrt{a}\times\sqrt{b}=\sqrt{ab}$

35. tomiko

now i get (attached)

36. asnaseer

that is correct - just one step left now - square both sides BTW: It is better to save your images as .png format if you can - they are smaller than .jpg files

37. tomiko

38. asnaseer

great! I'm glad you finally got there :)

39. tomiko

attached

40. asnaseer

perfect!

41. tomiko

sorry for all the bother. thanks for the help. i finished high school 6 years ago and didn't do any math at all and now just started uni. it's going to be a little change for me. i'm doing computer science now. lots of math in my class!! but i love subject.

42. asnaseer

no problem at all my friend - the main thing here is you obviously love to learn! :)

43. tomiko

challenge** thank you very much once again. now any question like this that comes my way i will murder it!!

44. asnaseer

he he - I love your "ambition" - keep up the good work! :)