Question:
Chapter:Real Numebrs. Video number : 5
In your 5th video, your proving that square of all positive integer is 3m (or) 3m 1...in that, concidering that if "r = 2", a = (3q + 2)^2 right... If we simplified it comes out as 9q^2 + 12q + 4 right ... Then, if we take 3 as common, it comes out as 3( 3q^2 + 4q ) + 4, This is how it should have been right ? But u proceeded like 3( 3q^2 + 4q + 1) + 1... Can you explain me why ?
Thank u in advance - Vish
Answer:
When you get
a = (3q + 2)2
=> a = 9q2 + 12q + 4
=> a = 9q2 + 12q + 3 + 1
=> a = 3(3q2 + 4q + 1) + 1
This is written in this form because we have to prove that square of all positive integers are either in the form of 3m or 3(m) + 1.
Here, we suppose m = 3q2 + 4q + 1
So a = 3m + 1.
Hope you understand now.
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