Mistakes give rise to Problems- 3
We know that a × ( b + c ) = ( a × b ) + ( a × c ) This property is there for "multiplication distributed over addition"
But if you do it in multiplication, then it's the following FALSE property a × ( b × c ) = ( a × b ) × ( a × c )
If you do it, it'll be a mistake !!!
But for how many distinct real values of a (independent of b and c , they can be anything except 0) is the above "False" property seen to be "always true" ?
The answer is 2.
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4 solutions
The answer should be 1 because it is given in the question that zero should not be considered...
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For b and c, not a.
0 and 1 are the values of a, not b and c
Exactly.....
suppose a,b,c=1 put this value in first 2-2=0 secondly 1-1=0
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Asked question was "number of values of a" and in your solution you already used a=1. b,c can be anything, a makes the difference.
wrong answer
wrong answer
+1 , -1
if we take 1 as a then there is no effect on the answer we can also take -1 therefore 1 and -1 r required ans
for a = 1 and -1 The false property is always true. That's why the answer is 2.
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Not for -1
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sir....can you kindly explain why "a" cannot be -1???????after all -1 also falls in the set of real distinct values!!!!!!!!!!
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@Debabrata Mukherjee – I am not sir, i am age 16. And see that if a=-1 , the LHS is − ( b × c ) and RHS is ( b × c ) and this is not true for ALL "b and c". Because we want independent of b and c, a must be eliminated and if a is -1, then what we get is not always true. − ( b × c ) = ( b × c ) . It's more about common sense dude!!! how will -1 make lhs=rhs for all?
@Debabrata Mukherjee – a cant be -1 because in that case abc != a^2bc for example take b=2 c=5 and a = -1 then a (b c)= -10 and (a b) (a*c)=10
-1 is not equal to 1 As -1^2 will be one
the only one number possible for this is 1
but a^2 =1
a=+1 and 0 (except 1/0)
Put a=1,2 and you get the answer. There cannot be any other value
I think you mean 1,0. Also, you have not provided the proof why there cannot be any other value.
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Proof is there 1 and 0 are those numbers which doesnt change its form.. b and c are real no.(R). R×1=R and R×0=0 while other real no.s would change the form....Hence it is PrOvEd...
simply the valus which will make equal effect on rhs and lhs even if they are multiplied several times.
two values are 1,0
As per the false property, we have a b c = a 2 b c and because b and c are non-zero, we get a = a 2 and thus the problem is actually finding the number of values of a such that a 2 = a and trivially, a 2 − a = 0 ⟹ a ( a − 1 ) = 0 ⟹ a = 0 or a = 1
Hence there are 2 values of a for which the false property is always true.