MA18A FAQ 1

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MATH 18 FAQ SESSION 1 | June 22, 2011 1

Answers to the questions are property of the Health Sciences Society

(Vistro-Yu, 1997)

Prove that if a , b > 0 then ab (a + b ) a 3 + b 3

SCRATCH WORK

Working backwards ,

ab(a + b) a 3 + b3

ab(a + b) (a + b) (a 2 – ab + b2) Recall that a 3 + b3 = (a + b) (a 2 – ab + b2).

0 (a + b) (a 2 – ab + b2) – ab(a + b) Transpose everything to the right hand side

0 (a + b) (a 2 – ab + b2 – ab) Distributive property

0 (a + b) (a 2 – 2ab + b2) Combining like terms

0 (a + b) (a – b)2 Recall that (a – b)2 = a 2 – 2ab + b2

Now we have to prove that the terms are either BOTH POSITIVE (or ZERO) or BOTH NEGATIVE (or

ZERO) so that the product is still POSITIVE (or ZERO).

Look at the second term (a – b)2. Recall that all square numbers are positive or zero. So this means that the

first term a + b must be POSITIVE or ZERO as well.

Recall the given statement a , b > 0. This means that a and b are both POSITIVE. Therefore, a + b > 0.

FORMAL PROOF

Since a , b > 0, it follows that a + b > 0.

Since (a – b)2 0 and a + b > 0, it follows that (a – b)2 (a + b) 0.

By distributive property,

(a

2

– 2ab + b

2

)

(a + b) 0(a + b) (a 2 – ab + b2) – ab(a + b) 0

Since a 3 + b3 = (a + b) (a 2 – ab + b2),

a 3 + b3 – ab(a + b) 0

a 3 + b3 ab(a + b) WHICH IS THE STATEMENT TO BE PROVEN.

(Ma18 Departmental Exam, 2007-2008 )

Let a , b be nonnegative numbers such that a b and a + b 1. Prove that 1 + 2ab 2a + b .

SCRATCH WORK

Working backwards ,

+ ³ +1 2 2ab a b

- ³ -2 2 1ab a b Transpose 2a to the left hand side, 1 to the right hand side

- ³ -2 ( 1) 1a b b Factoring the left hand side

- - - ³2 ( 1) ( 1) 0a b b Transpose b – 1 to the left hand side

(2 1)( 1) 0a b - - ³ By distributive property

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Now we have to prove that the terms are either BOTH POSITIVE (or ZERO) or BOTH NEGATIVE (or

ZERO) so that the product is still POSITIVE (or ZERO).

Notice that there is no OBVIOUS way of manipulating the given statements (a b and a + b 1) in such a

way that the final answer would be 2a – 1 or b – 1.

Let’s start first with the given statements.

Since a b, it follows that a – b 0.

Since a + b 1, it follows that a + b – 1 0.

Notice if we add the two statements a – b 0 and a + b – 1 0, we get:

a – b + a + b – 1 0

2a – 1 0 which is the first term!

Since 2a – 1 is NEGATIVE (or ZERO), then b – 1 must also be NEGATIVE (or ZERO).

To prove that b – 1 0, we go back to the original given statements.

Recall that a is nonnegative. It follows that a 0.

Since a + b 1, it follows that a 1 –

b.

Combining the two derived statements, we get 0 a 1 – b.

It follows that 0 1 – b. By multiplying both sides by –1, we get 0 b – 1.

We just proved that b – 1 is NEGATIVE (or ZERO)!

FORMAL PROOF

Since a b, it follows that a – b 0.

Since a + b 1, it follows that a + b – 1 0.

Since the sum of two negative numbers is negative, it follows thata – b + a + b – 1 0

2a – 1 0

Recall that a is nonnegative. It follows that a 0.

Since a + b 1, it follows that a 1 – b.

By transitivity, it follows that 0 1 – b OR 0 b – 1.

Since the product of two positive numbers (or zero) is positive (or zero), it follows that

- - ³(2 1)( 1) 0a b

By distributive property,

- - - ³2 ( 1) ( 1) 0a b b

- ³ -2 ( 1) 1a b b

- ³ -2 2 1ab a b

+ ³ +1 2 2ab a b

WHICH IS THE STATEMENT TO BE PROVEN.

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Prove: If a < b < 3, then b 2 + 6a – 2ab < 9.

SCRATCH WORK

Working backwards ,

+ - <2 6 2 9b a ab

+ - - <2 6 2 9 0b a ab Transpose 9 to the left hand side

- + - <2 9 6 2 0b a ab Rearranging the terms

- + - <2( 9) (6 2 ) 0b a ab Grouping the terms

+ - + - <( 3)( 3) 2 (3 ) 0b b a b Difference of two squares and factoring

Notice that both terms have a common factor of b – 3.

+ - - - <( 3)( 3) 2 ( 3) 0b b a b

( 3 2 )( 3) 0b a b + - - < By distributive property

The statement above states that the product of the two terms must be negative .

From the given b < 3, we can say that b – 3 < 0 (transpose 3 to the left hand side).

Since the second term b – 3 is NEGATIVE, the first term b + 3 – 2a must be POSITIVE.

To prove that, we go back to the original given statement a < b < 3.

There are three possible statements: a < 3, b < 3, and a < b.

Working backwards from our assumption b + 3 – 2a > 0,

b + 3 > 2a Transpose 2a to the right hand side

We will separate 2a to a + a . That way, the number of terms of both sides is equal and we can make a

comparison later on.b + 3 > a + a

Is b > a ? YES. Is 3 > a ? YES. Since both statements are true, it follows that b + 3 > a + a !

FORMAL PROOF

Since a < b < 3, it follows that: a < b, a < 3, b < 3

By addition property of inequality ,

a + a < b + 3

2a < b + 3

0 < b + 3 – 2a

Since b < 3, it follows that b – 3 < 0

Since b + 3 – 2a > 0 and b – 3 < 0, it follows that (b + 3 – 2a ) (b – 3) < 0.

By the distributive property,

b2 – 3b + 3b – 9 – 2ab + 6a < 0

b2 – 9 – 2ab + 6a < 0 – 3b + 3b will cancel each other

b2 – 2ab + 6a < 9 Transpose 9 to the right hand side


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Prove: If 1 3x - £ and 2 1y + £ , prove that 3 2 17y x - £ .

REMINDERS

1. Always start with the given statements.

2. Multiply the given with the factor corresponding to the coefficient of its variable in the final statement.

3. Using the absolute value term of the last statement, add/subtract the numbers corresponding to the

numerical constant of the variable in the absolute value term of the given statement after step 2.

4. Group the terms according to how the terms were expressed in step 2.

5. Proceed with the triangle inequality statement.

6. Replace all absolute value terms with the number corresponding to step 2.

7. Proceed with simplification and you should arrive at the final statement.

FORMAL PROOF- £1 3x + £2 1y

Multiply by 2 (by multiplication prop. of equality) Multiply by 3

- £2 1 3 2x + £3 2 1 3y

By the theorem =a b ab ,

- £2( 1) 6x + £3( 2) 3y

- £2 2 6x + £3 6 3y

- = - + - + - = + + - + -3 2 3 2 2 2 6 6 (3 6) ( 2 2) 8y x y x y x

By the triangle inequality theorem,

+ + - + - £ + + - - + -(3 6) ( 2 2) 8 3 6 2 2 8y x y x

+ + - + - £ + +(3 6) ( 2 2) 8 6 3 8y x

- £3 2 17y x


GENERAL TIPS FOR PROVING

1. Working backwards helps most of the time.

2. Experiment! You can transpose, factor, distribute, combine, etc. Do whatever it takes to make yourfinal statement look like the given.

3. Remember your basic theorems. Memorize them. It helps a lot.

4. Always have scratch work. It’s difficult if you don’t have one. You can show this in your answer sheet.

5. Learn to use logical statements (e.g. If/Since…, then/it follows that…)

6. As much as possible, provide a reason/theorem/property for every step you do.

7. Keep your formal proof neat!

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MORE EXERCISES


Let a and b be real numbers such that a < b < 1. Prove that a + ab < a 2 + b.

SCRATCH WORK

+ < +

- < -

- < -

- - <

2

2

( 1) ( 1)

( )( 1) 0

a ab a b

ab b a a

b a a a

b a a

Since a < b, it follows that b – a > 0. Since a < 1, it follows that a – 1 < 0.

FORMAL PROOF

Since a < b, it follows that b – a > 0. Since a < 1, it follows that a – 1 < 0.

Since the product of a negative and a positive number is negative, it follows that

- - <( )( 1) 0b a a

- - - <( 1) ( 1) 0b a a a Distributive property

- < -( 1) ( 1)b a a a

- < -2ab b a a Distributive property

+ < +2a ab a b



Let a , b , c , d be positive numbers. Prove that (a + b ) (b + c ) (c + d ) (d + a ) 16abcd .

FORMAL PROOF

By arithmetic mean-geometric mean theorem (AM-GM),

+³

2

a bab

+³

2

b c bc

+³

2

c d cd

+³

2

d a da .

By multiplication property of inequality,

( )( )( )( )æ öæ öæ öæ ö+ + + +÷ ÷ ÷ ÷ç ç ç ç ³÷ ÷ ÷ ÷ç ç ç ç÷ ÷ ÷ ÷ç ç ç ç÷ ÷ ÷ ÷è øè øè øè ø2 2 2 2

a b b c c d d a ab bc cd da

+ + + +³ 2 2 2 2( )( )( )( )

16

a b b c c d d a a b c d

+ + + +³

( )( )( )( )

16

a b b c c d d a abcd

(a + b) (b + c ) (c + d ) (d + a ) 16abcd


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(Ma18 Departmental Exam, 2003-2004)

Let a and b be real numbers with a > b > 0. Prove that a (2a + b ) > b 2.

SCRATCH WORK

+ >

+ >

+ - >

- + >

2

2 2

2 2

(2 )

2

2 0

(2 )( ) 0

a a b b

a ab b

a ab b

a b a b

Since a > b and a > 0, it follows that 2a > b. It also follows that 2a – b > 0.

Since a > 0 and b > 0, it follows that a + b > 0.

FORMAL PROOF

Since a > b and a > 0, it follows that 2a > b. It also follows that 2a – b > 0.

Since a > 0 and b > 0, it follows that a + b > 0.Since the product of two positive numbers is positive,

- + >(2 )( ) 0a b a b

+ - - >2 22 2 0a ab ab b Distributive property

+ - >2 22 0a ab b

+ >2 22a ab b

+ <2(2 )a a b b WHICH IS THE STATEMENT TO BE PROVEN.

(Rivera, 1999 )

Prove that if x > y > 0 then 4x 3 (x – y ) > x 4 – y 4

SCRATCH WORK

- > -

- > - +

- > - +

- > - +

- > - + +

- > - + + +

3 4 4

3 2 2 2 2

3 2 2 2 2

3 2 2 2 2

3 2 2

3 3 2 2 3

4 ( )

4 ( ) ( )( )

4 ( ) ( )( )

4 ( ) ( )( )

4 ( ) ( )( )( )

4 ( ) ( )( )

x x y x y

x x y x y x y

x x y x y x y

x x y x y x y

x x y x y x y x y

x x y x y x xy x y y

Since x > y > 0, x – y > 0

Dividing both sides by x – y ,

> + + +3 3 2 2 34x x xy x y y

We can prove the statement above by making the number of terms equal for both sides. We can make the left

hand side four terms by expanding 4x 3.

+ + + > + + +3 3 3 3 3 2 2 3x x x x x xy x y y

Since x > y , it follows that: x 3 > xy 2, x 2y , y 3.

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FORMAL PROOF

Since x > y > 0, by the multiplication property of inequality, it follows that:

x 3 > xy 2

x 3 > x 2y

x 3 > y 3

By addition property of inequality,

x 3 + x 3 + x 3 > xy 2 + x 2y + y 3

3x 3 > xy 2 + x 2y + y 3

Adding x 3 to both sides,

3x 3 + x 3 > x 3 + xy 2 + x 2y + y 3

4x 3 > x 3 + xy 2 + x 2y + y 3

Factoring the right hand side,

4x 3 > (x + y ) (x 2 + y 2)

Since x > y , it follows that x – y > 0.Multiplying x – y to both sides,

4x 3 (x – y ) > (x – y ) (x + y ) (x 2 + y 2)

4x 3 (x – y ) > (x 2 – y 2) (x 2 + y 2)

4x 3 (x – y ) > x 4 – y 4


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School Year 2011-2012

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