1:-)
Prove (-1) x (-1) = 1 algebraically ....
Soln. can be given by anyone other than ATGS :-)
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14 Answers
49eeeeeeeeeeeeeeeeeeee..........dats not fair.........bt okkies...........
well, try to prove it by other ways too.....i already hav two solns for it.....
24Read it somewhere..
"""It is kind of like two wrongs make a right. Let us say that x and
y are both positive. Then the meaning of -(x) + (y) is just y - x and (y) - -(x) = y + x, subtracting a negative number is the same as adding the positivenumber. Now, let us say n is a positive integer. Then n * x can be thought of as adding x to itself n times which explains why a positive number times a negative number is always negative. So, what does -(n) * x?
Well,multiplication is commutative, i.e., x * y = y * x. So -(n) * x = x * -(n)
or -(n) * x = (-1 * n ) * x = ( n * -1 ) * x = n * ( -1 * x ) becausemultiplication is also associative = n * -(x). All of which is to say that -n* x is the same as adding -x to itself n times. Therefore, -n * -x is the sameas adding -(-x) to itself n times. And, I think we all agree that -(-x)better be +x. """
49bapre bap itna lamba proof.........it is much easy than this.....
1i would like to say
like this
1x1=1 implies addin 1 one time
-1x1=-1 i.e addin -1 one time
1x-1 =-1 addin 1 , minus one time i.e subtractin one, 1 times
-1x-1=1 addin -1,minus one times that is
subtractin -1 ,-1 times
62(-1)(1-1) = 0
=> (-1).1 +(-1)(-1) = 0
=> -1 +(-1)(-1) = 0
=> (-1)(-1) = 1
24DOnt know if this is right or wrong.
(-1)=e^{-i\pi}
So
(-1).(-1)=e^{-i\pi}.e^{-i\pi}=>(-1).(-1)=e^{-2i \pi}=1
341that wouldnt be right. You have to establish this property on the basis of the properties of integers as a division ring with unit element (Division ring is an algebraic structure. wiki it)
To be rigorous, we first have to establish a X 0 = 0
Proof: a X 0 = a X (0+0) (0 is the Additive Identity)
= aX0 + aX0 (Distributive law of multiplication over addition)
So, if a X 0 = c for some integer c (c will be an integer from closure property), then we have c = c + c.
Hence 0 = c + (-c) = c + c+(-c) = c + [c+(-c)] (Associative law for addition]
or 0 =c+0 = c
Thus a X 0 = 0
Using this we have
1 + (-1) = 0 (Additive inverse)
Hence 0 = (-1)X0 = (-1) X [1+(-1)] = (-1) X 1 + (-1) X (-1)
= -1 + (-1) X (-1) [1 is the multiplicative identity]
Thus 1 = 1+0 = 1+(-1) + (-1) X (-1) = 0 + (-1) X (-1) = (-1) X (-1)
DeMoivre's Theorem would need you to assume the field properties of real numbers which is a bigger assumption than what you need here.
62prophet sir.. so you have studied rings :D
btw.. Did you see the proof of chinese remainder theorem by algebra.. (one proof that i love very much :)
btw if we start talking about those.. this forum will be one for BSc students :P
49well this was one of my first googlies....i used trigo to prove this..........
let -1=cosθ
nw cosθ=1
so, θ=π
so, sinθ=0
so (-1)(-1)=cos2θ=1-sin2θ=1-0=1
hence proved[9][9][9]