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For benefit of all people I am writing the proof of the 5 mark (9+45) Question.

(9+45)^48 = + f.

(9+45)^48 (9-45)^48 = 1.

( + f) (9-45)^48 = 1.

x(9-45)^48 = 1.

So we have to prove that (9-45)^48 = 1-f so that x(1-f) =1.

Now, (9+45) ^48 + (9-45) ^48 = 2(9^48 + __) = Even Integer.

(9-45)^48 < 1 since (9-45) < 1. Let (9-45) = f'. 0 < f' < 1.

Hence + f + f' = Even Integer.

Since 0<1 and 0< f' <1. f + f' =1. Hence f' = 1-f.

QED.

(9+45)^48 = + f.

(9+45)^48 (9-45)^48 = 1.

( + f) (9-45)^48 = 1.

x(9-45)^48 = 1.

So we have to prove that (9-45)^48 = 1-f so that x(1-f) =1.

Now, (9+45) ^48 + (9-45) ^48 = 2(9^48 + __) = Even Integer.

(9-45)^48 < 1 since (9-45) < 1. Let (9-45) = f'. 0 < f' < 1.

Hence + f + f' = Even Integer.

Since 0

QED.