Math Q (1 Viewer)

[axwe7]

Need help with (a) and (b),

Cheers,
Axwe7.

 

[leehuan]

Just having a wild stab at a):

Equating expressions under the surd - y+2=5 ----> y=3
Equating expressions that do not get taken indices of - x-3 = -1 ----> x=2

b) would be done similarly

 

[axwe7]

Just having a wild stab at a):

Equating expressions under the surd - y+2=5 ----> y=3
Equating expressions that do not get taken indices of - x-3 = -1 ----> x=2

b) would be done similarly

Yep, that's right!
TBH, I never knew it's that simple...
However, are you sure that that is the right method?

 

[glittergal96]

However, are you sure that that is the right method?

To justify the "equating like parts", use the following result.

If a,b,c,d are rationals, c,d >= 0 and c is not the square of a rational, then

a+sqrt(c)=b+sqrt(d)

implies a=b and c=d.

Proof:

Note that if we either have a=b or c=d that the other equality falls out immediately, so let us assume that neither of these inequalities hold.

Then a-b=sqrt(d)-sqrt(c),

=> (a-b)(sqrt(d)+sqrt(c))=d-c

=> sqrt(d)+sqrt(c)=(d-c)/(a-b) is rational.

As sqrt(d)-sqrt(c) is also rational, this allows us to deduce that sqrt(c)=(sqrt(d)+sqrt(c))-(sqrt(d)-sqrt(c))/2 is rational, contrary to our assumptions.

Hence a=b and c=d.

 

[glittergal96]

^ This result is the rigorous reasoning that solves pretty much all of the problems you have posted today, so make sure you understand it.

 

[axwe7]

^ This result is the rigorous reasoning that solves pretty much all of the problems you have posted today, so make sure you understand it.

Thanks so much for all of that, it's just that, the cream behind the sugar coating was a little bit foggy. I just didn't understand the basics.

Thanks to you, now I have.

Thanks so much.

Cheers,
Axwe7