Answer to Question #284929 in Algebra for Bjoy

Question

 Use inductive reasoning to make a conjecture about a rule that relates the number you selected to final answer. Try to prove your conjecture by using deductive reasoning.

1. Pick a number

Double it

Subtract 20 from the answer

Divide by 2

Subtract the original number Result:

2. Pick a number Multiply it by 9

Add 21

Divide by 3

Subtract three times the original number Result:

3. Pick a number

Add 6

Multiply the answer by 9

Divide the answer by 3

Subtract 3 times the original number Result

Answer 1

1.

Let the number be $1.$

Double it

$$2\cdot1=2$$

Subtract 20 from the answer

$$2-20=-18$$

Divide by 2

$$\dfrac{-18}{2}=-9$$

Subtract the original number

$$-9-1=-10$$

Result:

$$-10$$

Let the number be $-\dfrac{1}{3}.$

Double it

$$2\cdot(-\dfrac{1}{3})=-\dfrac{2}{3}$$

Subtract 20 from the answer

$$-\dfrac{2}{3}-20=-\dfrac{62}{3}$$

Divide by 2

$$\dfrac{-\dfrac{62}{3}}{2}=-\dfrac{31}{3}$$

Subtract the original number

$$-\dfrac{31}{3}-(-\dfrac{1}{3})=-10$$

Result:

$$-10$$

Let the number be $\sqrt{2}.$

Double it

$$2\cdot(\sqrt{2})=2\sqrt{2}$$

Subtract 20 from the answer

$$2\sqrt{2}-20=2\sqrt{2}-20$$

Divide by 2

$$\dfrac{2\sqrt{2}-20}{2}=\sqrt{2}-10$$

Subtract the original number

$$\sqrt{2}-10-\sqrt{2}=-10$$

Result:

$$-10$$

A conjecture: the result is always $-10.$

Take arbitrary number $c\in\R.$

Double it

$$2\cdot(c)=2c$$

Subtract 20 from the answer

$$2c-20=2c-20$$

Divide by 2

$$\dfrac{2c-20}{2}=c-10$$

Subtract the original number

$$c-10-c=-10$$

Result:

$$-10$$

We see that the result does not depend on the taken number.

Therefore we proved that the result is always $-10$ regardless of the number we pick.

2.

Let the number be $1.$

Multiply it by 9

$$1\cdot9=9$$

Add 21

$$9+21=30$$

Divide by 3

$$\dfrac{30}{3}=10$$

Subtract three times the original number

$$10-3(1)=7$$

Result:

$$7$$

Let the number be $\dfrac{1}{2}.$

Multiply it by 9

$$\dfrac{1}{2}\cdot9=\dfrac{9}{2}$$

Add 21

$$\dfrac{9}{2}+21=\dfrac{51}{2}$$

Divide by 3

$$\dfrac{\dfrac{51}{2}}{3}=\dfrac{17}{2}$$

Subtract three times the original number

$$\dfrac{17}{2}-3(\dfrac{1}{2})=7$$

Result:

$$7$$

Let the number be $\sqrt{2}.$

Multiply it by 9

$$\sqrt{2}\cdot9=9\sqrt{2}$$

Add 21

$$9\sqrt{2}+21=9\sqrt{2}+21$$

Divide by 3

$$\dfrac{9\sqrt{2}+21}{3}=3\sqrt{2}+7$$

Subtract three times the original number

$$3\sqrt{2}+7-3(\sqrt{2})=7$$

Result:

$$7$$

A conjecture: the result is always $7.$

Take arbitrary number $c\in\R.$

Multiply it by 9

$$c\cdot9=9c$$

Add 21

$$9c+21=9c+21$$

Divide by 3

$$\dfrac{9c+21}{3}=3c+7$$

Subtract three times the original number

$$3c+7-3c=7$$

Result:

$$7$$

We see that the result does not depend on the taken number.

Therefore we proved that the result is always $7$ regardless of number we pick.

3.

Let the number be $1.$

Add 6

$$1+6=7$$

Multiply the answer by 9

$$7\cdot9=63$$

Divide the answer by 3

$$\dfrac{63}{3}=21$$

Subtract 3 times the original number

$$21-3(1)=18$$

Result:

$$18$$

Let the number be $-\dfrac{1}{2}.$

Add 6

$$-\dfrac{1}{2}+6=\dfrac{11}{2}$$

Multiply the answer by 9

$$\dfrac{11}{2}\cdot9=\dfrac{99}{2}$$

Divide the answer by 3

$$\dfrac{\dfrac{99}{2}}{3}=\dfrac{33}{2}$$

Subtract 3 times the original number

$$\dfrac{33}{2}-3(-\dfrac{1}{2})=18$$

Result:

$$18$$

Let the number is $\sqrt{2}.$

Add 6

$$\sqrt{2}+6=\sqrt{2}+6$$

Multiply the answer by 9

$$(\sqrt{2}+6)\cdot9=9\sqrt{2}+54$$

Divide the answer by 3

$$\dfrac{9\sqrt{2}+54}{3}=3\sqrt{2}+18$$

Subtract 3 times the original number

$$3\sqrt{2}+18-3(\sqrt{2})=18$$

Result:

$$18$$

A conjecture: the result is always $18.$

Take arbitrary number $c\in\R.$

Add 6

$$c+6=c+6$$

Multiply the answer by 9

$$(c+6)\cdot9=9c+54$$

Divide the answer by 3

$$\dfrac{9c+54}{3}=3c+18$$

Subtract 3 times the original number

$$3c+18-3(c)=18$$

Result:

$$18$$

We see that the result does not depend on the taken number.

Therefore we proved that the result is always $18$ regardless of number we pick.