Can you explain it to me.
2006-10-23 02:44:06 · 4 answers · asked by Rhonda O 1 in Science & Mathematics ➔ Mathematics
It is true for nonnegative real numbers, yes. In fact, if you allow imaginary numbers, then it is true for negative reals as well. This is because while x^2 = a has two real solutions when a is a positive real, sqrt(a) specifically refers to the positive real solution. If a is a negative real, sqrt(a) refers to the multiple of i (the imaginary number) with a positive coefficient. It is by definition true that squaring the square root yields the original number.
2006-10-23 02:51:38 · answer #1 · answered by DavidK93 7 · 3⤊ 1⤋
it is only an identity if it is true for all numbers, in the set you are looking at. if you are just looking at nonnegative values then its fine, but if you do not impose this restriction then its not an identity since its not satisfied for negative values in the real domain
2006-10-23 10:41:10 · answer #2 · answered by tsunamijon 4 · 0⤊ 1⤋
a square root basically is a power of 1/2. So it's really x^1/2. So when you square that you get (x^1/2)^2, you take the 1/2 and multiply it by the 2. So you get x^(2/2). Does that make sense?
2006-10-23 09:50:57 · answer #3 · answered by The C man 3 · 0⤊ 1⤋
it is false when x <0,
the first thing to notice is that if x < 0 then
you cannot take its square root, i.e, âx is NOT defined
2006-10-23 11:20:15 · answer #4 · answered by locuaz 7 · 0⤊ 0⤋