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Multiplying a number with a fraction

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2

I was writing some converters of units using BigDecimals and I ran across a situation where I had to multiply a number with a fraction - periodic number.

For most cases the precision is good enough, but lets say we have an equation like:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10)
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP))

Normally this would equal 10, however because of rounding, we will get 9.999999...

Is there an elegant way of implementing this without having an if condition detecting when the fraction can be cut?

java math floating-point division

share|improve this question

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you want to deal correctly with fractions, you'll need a Fraction class of some kind. Either roll your own, or have a look at what Apache provides.
  
  – Dawood ibn Kareem
  Mar 24 at 21:10
  

  
  
  
  

add a comment | 

2

I was writing some converters of units using BigDecimals and I ran across a situation where I had to multiply a number with a fraction - periodic number.

For most cases the precision is good enough, but lets say we have an equation like:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10)
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP))

Normally this would equal 10, however because of rounding, we will get 9.999999...

Is there an elegant way of implementing this without having an if condition detecting when the fraction can be cut?

java math floating-point division

share|improve this question

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you want to deal correctly with fractions, you'll need a Fraction class of some kind. Either roll your own, or have a look at what Apache provides.
  
  – Dawood ibn Kareem
  Mar 24 at 21:10
  

  
  
  
  

add a comment | 

I was writing some converters of units using BigDecimals and I ran across a situation where I had to multiply a number with a fraction - periodic number.

For most cases the precision is good enough, but lets say we have an equation like:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10)
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP))

Normally this would equal 10, however because of rounding, we will get 9.999999...

Is there an elegant way of implementing this without having an if condition detecting when the fraction can be cut?

java math floating-point division

share|improve this question

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10)
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP))

share|improve this question

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

share|improve this question

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

edited Mar 26 at 20:23

Andronicus

7,54432035

edited Mar 26 at 20:23

Andronicus

7,54432035

asked Mar 24 at 20:27

RadovanRadovan

77110

asked Mar 24 at 20:27

RadovanRadovan

77110

2 Answers
2

active

oldest

votes

3

The following will work:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP)

The difference is that here the operations are chained, which allows for resolving such cases. In your solution the division needs to be calculated (where error occurs) and then multiplication, because it's passed as argument.

share|improve this answer

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Awesome. Normally I prefer having parenthesis everywhere when it comes to math in code. Just makes reading and reasoning about the order of operations a lot easier. Didn't realize it interferes with precision. Thanks a lot!
  
  – Radovan
  Mar 25 at 9:38
  

  
  
  
  

add a comment | 

1

Do not know if this will be a general case answer for you, but it works in the example given:

bd = BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90));

Multiply by 10 then divide by 90.

a * x = ax
 - --
 z z

You will need to include some rounding logic for rational numbers:

bd = BigDecimal.valueOf(1)
 .multiply(BigDecimal.valueOf(1))
 .divide(BigDecimal.valueOf(3));

Will fail without rounding.

share|improve this answer

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

add a comment | 

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3

The following will work:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP)

The difference is that here the operations are chained, which allows for resolving such cases. In your solution the division needs to be calculated (where error occurs) and then multiplication, because it's passed as argument.

share|improve this answer

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Awesome. Normally I prefer having parenthesis everywhere when it comes to math in code. Just makes reading and reasoning about the order of operations a lot easier. Didn't realize it interferes with precision. Thanks a lot!
  
  – Radovan
  Mar 25 at 9:38
  

  
  
  
  

add a comment | 

3

The following will work:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP)

The difference is that here the operations are chained, which allows for resolving such cases. In your solution the division needs to be calculated (where error occurs) and then multiplication, because it's passed as argument.

share|improve this answer

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Awesome. Normally I prefer having parenthesis everywhere when it comes to math in code. Just makes reading and reasoning about the order of operations a lot easier. Didn't realize it interferes with precision. Thanks a lot!
  
  – Radovan
  Mar 25 at 9:38
  

  
  
  
  

add a comment | 

The following will work:

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP)

The difference is that here the operations are chained, which allows for resolving such cases. In your solution the division needs to be calculated (where error occurs) and then multiplication, because it's passed as argument.

share|improve this answer

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90), 6, RoundingMode.HALF_UP)

share|improve this answer

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

answered Mar 24 at 20:39

AndronicusAndronicus

7,54432035

1

Do not know if this will be a general case answer for you, but it works in the example given:

bd = BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90));

Multiply by 10 then divide by 90.

a * x = ax
 - --
 z z

You will need to include some rounding logic for rational numbers:

bd = BigDecimal.valueOf(1)
 .multiply(BigDecimal.valueOf(1))
 .divide(BigDecimal.valueOf(3));

Will fail without rounding.

share|improve this answer

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

add a comment | 

1

Do not know if this will be a general case answer for you, but it works in the example given:

bd = BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90));

Multiply by 10 then divide by 90.

a * x = ax
 - --
 z z

You will need to include some rounding logic for rational numbers:

bd = BigDecimal.valueOf(1)
 .multiply(BigDecimal.valueOf(1))
 .divide(BigDecimal.valueOf(3));

Will fail without rounding.

share|improve this answer

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

add a comment | 

Do not know if this will be a general case answer for you, but it works in the example given:

bd = BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90));

Multiply by 10 then divide by 90.

a * x = ax
 - --
 z z

You will need to include some rounding logic for rational numbers:

bd = BigDecimal.valueOf(1)
 .multiply(BigDecimal.valueOf(1))
 .divide(BigDecimal.valueOf(3));

Will fail without rounding.

share|improve this answer

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

bd = BigDecimal.valueOf(90)
 .multiply(BigDecimal.valueOf(10))
 .divide(BigDecimal.valueOf(90));

a * x = ax
 - --
 z z

bd = BigDecimal.valueOf(1)
 .multiply(BigDecimal.valueOf(1))
 .divide(BigDecimal.valueOf(3));

share|improve this answer

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112

answered Mar 24 at 20:31

Not a JDNot a JD

1,367112