help, math hw?

Question

heres the problem

a set of numbers is said to be closed under a certain operation if, when you perfrom the operation any 2 #s in a set, the result is also in a set.

the set of real #s is closed under addition. This means that when you add any two real numbers, the sum is also a real #

is the set of real numbers closed under subtraction?

Answer 1

The set of real numbers is closed under subtraction because the subtraction involving two real numbers also results in a real number.

Answer 2

i will attempt to describe the terrific i will: before everything, whenever you opt to unravel for a variable (consequently x), you ought to placed the variable on one area and the different numbers on the different. So first placed the x on the left area of the equation. to try this, you would be conscious that x is destructive (-x). with the aid of this, to stability the equation, you upload effective x to the two area: 5 + x = 8 - x + x Now, you already know that -x + x equals 0, so which you will now take that out of the equation: 5 + x = 8 Now you ought to placed the 5 on the wonderful area so x could be on my own. you already know that 5 is an outstanding form (+5), so which you ought to stability the equation, you subtract 5 on the two area: 5 - 5 + x = 8 - 5 5 - 5 equals 0, so which you will now take that out of the equation: x = 8 - 5 you are able to now resolve x surely. x = 3 So this implies the appropriate answer is x = 3 If we plug in 3 to the equation, we are going to se that this is genuine. 5 = 8 - 3 5 = 5 i'm hoping this facilitates =)

Answer 3

If you subtract one number from another number, do you get a real number as the result?

If you suspect the answer is "yes", then R is closed under subtraction.

If you suspect the answer is "no", then provide a counterexample.