A Problem of Mathematics

[s_m_u_yousaf]

Aslaam-o-Alaikum

Is below statement true? Plz answer with examples.

( √Any Number)2 = √(Any Number)2

√ = Under-Root

[Enlightened]

If "any number" on both sides is the same number. For example (sqr(4))2=(sqr(4))2. In other words if you remove multiplier of 2 from both sides, you have only sqr(any number) left. Thus (any number) has to be the same number on both sides or we can say any number to be "x" then "x" has to be on both sides in above example given as 4. I think in math it will make sense but in plain english it might confuse some people. I hope it makes sense.

(sqr(x))2= (sqr(x))2 is equal to

(sqr (x)) = (sqr (x))

where "x" is any number.

[s_m_u_yousaf]

@ Enlightened
U tried well with given data but Let me make my Question more clear

( √x)2 = √(x)2

√ = Under-Root
x belongs to "Integers"

Plz try it again with solid arguments

Actually 2 means Square
Mujh se wo Super Script main ja nahein raha thaa
Mra Khayaal ahi k ab sari situation clear ho gayi ho gi
So try it again

[Enlightened]

@ yousaf,

Seems like I misunderstood the question as it was not very clear.

Anyways, let me try again as I understand it.

The question is whether (sqrt(x))^2 = sqrt (x^2)

So dear, we can also write this as (x^1/2)^2 = (x^2)^1/2

now, if you further manipulate the equation using mathematical laws then equation becomes:

x^2/2 = x ^ 2/2 and further it becomes x^1 = x ^1 and finally we have

x=x

I hope it makes sense. Please correct me if I am wrong as it has been a very long time since I used these laws.

[s_m_u_yousaf]

Aslaam-o-Alaikum
@ Enlightened
Janaab now I tell you one thing more. Please check it.
In mathematics A statement is true only when it fulfills all conditions
I think you did not give attention to information about x
x belongs to Intergers.
Integers can be positive or negative.
Statement given by me is true for all Positive integers, but, for negative numbers it is not always true. Now I will give you a way solving the statement if I am at mistake then plz rectify the error.
Possible Solution:
Let us Suppose
x = -3
L.H.S:-
= (-3^1/2)^2
= (((3)(-1))^1/2)^2
= ((3)^1/2 (-1)^1/2)^2
= ((3)^1/2 . i)^2 Where i = Iota
= ((3)^1/2)^2 . (i)^2
= 3. (-1) i^2 = -1
= -3
R.H.S:-
= ((-3)^2)^1/2
= (9)^1/2
= ((3)^2)^1/2
= 3
So,
L.H.S. is not equal to R.H.S
What do you think about it?

[Enlightened]

Thanks,

Actually, I am aware of each and every rule you mentioned including complex numbers. I just did not evaluate the negative aspect of the equation as hasty as I was to answer the question. Believe it or not, I have that problem of overlooking information even when on tests.

However let me point out something about the L.H.S last lines.
"= 3. (-1) i^2 = -1
= -3"

I think i^2 = -1, Would not 3. (-1).i^2 give you
=3.(-1).(-1) = 3,

I think you just made a typo. Thanks for the mind exercise.

[WaqasAhmed]

in problem, on RHS we are squaring the no. first and then taking under root which may result a negative no. as well. so both sides cant be equal. but taking modulus will alway yeild same no. on both sides.

[tanolig]

Have a look here.
Let Any number=x=4
then
L.H.S=(sqr(4))*2=2*2=4; ------->(i)
But
R.H.S=sqr(4*2)=sqr(8)=2*sqr(2)=2.8284; -------->(ii)
From (i) and (ii) it is clear that L.H.S=/=R.H.S. As statement is not true for x=4. Hence it is not true in general. Best Regards!

[Ayesha Mahmood]

Originally Posted by tanolig

Have a look here.
Let Any number=x=4
then
L.H.S=(sqr(4))*2=2*2=4; ------->(i)
But
R.H.S=sqr(4*2)=sqr(8)=2*sqr(2)=2.8284; -------->(ii)
From (i) and (ii) it is clear that L.H.S=/=R.H.S. As statement is not true for x=4. Hence it is not true in general. Best Regards!

4^2 = 16 not 8

[tanolig]

Dear Ayesha Mahmood,
Question was ( √Any Number)2 = √(Any Number)2
Look at right side of the above problem. On the R.H.S binary opration used between 'Any Number' and '2' within the square root is multiplication, not a power opration.
If it is a power opration then (√(Any Number)^2)=Any Number