Have you tried expanding the expression?
5 × 8 7 × 1 2 1 0 × 1 7 is larger than ( 5 − 1 ) × ( 8 + 1 ) is larger than ( 7 − 1 ) × ( 1 2 + 1 ) is larger than ( 1 0 − 1 ) × ( 1 7 + 1 )
Is it true that for all positive integers A and B ( with A < B ) , the value of A × B must be larger than ( A − 1 ) × ( B + 1 ) ?
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3 solutions
Here, ( A − 1 ) ( B + 1 ) = A B + A − B − 1
( A − 1 ) ( B + 1 ) = A B + A − B − 1 > A B if A − B − 1 < 0
Simplifying, A < B + 1 .
Now we can say, ( A − 1 ) ( B + 1 ) > A B if A < B + 1 . But given that, A < B which is A < B < B + 1 , always true for all positive integer.
So, (boxed{\text{It's true.}}
True if A,B>1 Incase A=1 B=2 then this equation does not hold.
The answer is yes lol
By foiling out the variables we can expand the given expression: ( A − 1 ) ( B + 1 ) = A B + A − B − 1
As it is given that B > A , we can rewrite our expression as A B − k − 1 , where k is a positive integer. (We replaced A − B with − k )
A B − k − 1 will always be less than A B , given that A , B , k are positive integers