Why do negative numbers have no square root?

You can think backwards
That is, can we get a negative number after two negative numbers are squared? We know that negative * negative = positive
So negative numbers can't have square roots

RELATED INFORMATIONS

1. Why does the value of root 6 have no negative number, only positive number, but the square root of 6 has negative number?
2. Do negative numbers have square roots? Is an imaginary number the square root of a negative number I remember that negative numbers have no square root What is the imaginary number? If not, why not
3. Is the negative 2 square of a negative number a positive number, such as (- 2) ^ - 2
4. How to calculate the square of negative number
5. How to calculate the negative square of a constant
6. How to calculate the negative square of a number? No, it's a power. It's a negative number
7. Is the square of a negative number negative or positive without brackets?
8. It is known that: x2 + Y2 + 4x-6y + 13 = 0, where x and y are rational numbers, then YX=______ ．
9. Given that x ^ 2 + y ^ 2 + 4x-6y + 13 = 0, X and y are rational numbers, find the value of x ^ y
10. Write 5 rational numbers (which cannot be repeated), 3 of which are non negative; 3 of which are non positive; 5 of which are all positive
11. Can't the root of the root be negative and zero?
12. Simple operation of the power of the product of grade seven 1) The fifth power of 2 × the fourth power of 5 2) (- two thirds) to the power of 2005 × 1.5 to the power of 2005 3) 8 to the power of 8 × (- 1 / 4) to the power of 12 4) The fifth power of (- 3) × (- 2 / 3) × the sixth power of (- 5) It's gone Don't give me a short answer - thank you first.
13. The power of 3-trace product (simple operation + process) (1)2^5*5^4 (2)8^8*(-1/4)^12 (3)(-3)^5*(-2/3)^5*(-5)^6 For these three questions, it is necessary to simplify the calculation and complete the process! 2^5x5^4 8^8x(-1/4)^12 (-3)^5x(-2/3)^5x(-5)^6
14. Simple calculation of product's power 1.8^3×（1/4)^4 2.(-1/8)^3×2^9
15. Part eight: the calculation of product's power, 1.[(x-y^3]^4*[-(y-x)^2]^5*(x-y) 2.(-2x^2y)^3-8(x^2)^2*(-x)^2*(-y)^3 The second question is wrong 2.(-2x^2*y)^3-8(x^2)^2*(-x)^2*(-y)^3 One more question 3.(a^m+b^n-2)* (-2a^m-2b^n-2)^2* (3a^m+3b^n-2)^3* (1/2b^n-2+1/2a^m)^4
16. On the calculation of product's power ② （-a^3*b^6）^4+（-a^4*b^8）^3 ③ -（-x^2）^3*（-x^2）^2-x*（-x^3）^3 ④（-2a^2）^3 + （-3a^3）^2 +（-a）^6
17. A problem about the power of product is going to be solved tonight (0.5 times 3 and 2 / 3) 2009 times (- 2 times 3 / 11) 2008
18. Several problems about the power of product 1. The third power of ab 2. The fourth power of (- AB) 3. (x to the power of 3, y to the power of 2m) 4. The 5th power of (2m power of x) - the 2m power of 5 (5m power of x) 5. The second power of (- 3x) and the third power of [(- 2x)]
19. The formula of product's power calculation What is the sum of the sixth power of (- 2XY quadratic) and the third power of (- 3x quadratic y quartic)
20. If x + 1 power of 3 times x power of 2 - x power of 3 times x + 1 power of 2 = 2 power of 2 times 2 power of 3 Find x?