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The set of rational numbers is closed under the basic operations of arithmetic.

Suppose 3 + 2x5 is rational.

Subtract 3, which is a rational number: then 2x5 is rational

Divide by 2, a rational number: then x5 is rational.

But you are given that x5 is irrational therefore the supposition was wrong. That is, 3 + 2x5 is not rational.

Q: How can you prove that 3 plus 2x⁵ is an irrational number where x⁵ is irrational?

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This can't be proved because it is not necessarily true. If x is 2 and y is 1 then x-y is 1, which is rational. The product of 2x and 3y is 12, which is also rational. Sadly, you can't disprove it either; for certain values of x,y it is true as posited.

2x + 27, where x is your number

2X+20 Where "X" is your number.

(Y = -2x plus or minus any number) is parallel to (Y = -2x + 5) .

2x+4y

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This can't be proved because it is not necessarily true. If x is 2 and y is 1 then x-y is 1, which is rational. The product of 2x and 3y is 12, which is also rational. Sadly, you can't disprove it either; for certain values of x,y it is true as posited.

2x + 27, where x is your number

[ y = 2x plus or minus any number ] is parallel to it. [ y = -0.5x plus or minus any number ] is perpendicular to it.

2X+20 Where "X" is your number.

Twice a number can be represented as 2x, where x is the unknown number. Adding 27 to this expression gives 2x + 27. Therefore, "twice a number plus 27" can be written as the algebraic expression 2x + 27.

2x + 27

8x-4

[ y = -2x + any other number ] is parallel to [ y = -2x + 6 ].

(Y = -2x plus or minus any number) is parallel to (Y = -2x + 5) .

3x + 2x - 8x + 2x = -x

2x+4y

x=1