A problem by Nazmus Sakib
True or False:
1 2 + 1 2 1 2 4 + 2 1 2 1 ∈ Z
Note: By breaking , if you want to change the sign then the sign of the numerator and denominator should be the same. e.g c + d 2 a + b 2 = c + d a + b or c − d a − b
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4 solutions
1 2 + 1 2 1 2 4 + 2 1 2 1
= 1 2 + 1 1 2 4 + 2 × 1 1
= 2 3 4 6
= 2 ∈ Z
1 2 + 1 2 1 2 4 + 2 1 2 1 = 1 2 + 1 2 1 2 ( 1 2 + 1 2 1 ) = 2
1 2 + 1 2 1 2 4 + 2 1 2 1 ⇒ 1 ( 1 2 + 1 2 1 ) 2 ( 1 2 + 1 2 1 ) ⇒ 2
1 2 + 1 2 1 2 4 + 2 1 2 1
= 1 2 + 1 1 2 4 + 2 2 ;(with plus sign)
= 2 ( integer)
or, 1 2 − 1 1 2 4 − 2 2 ;(with minus sign)
= 2 ( integer)