Convert -6.800 000 190 734 863 287 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

-6.800 000 190 734 863 287(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. Start with the positive version of the number:

|-6.800 000 190 734 863 287| = 6.800 000 190 734 863 287

2. First, convert to the binary (base 2) the integer part: 6.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.

  • division = quotient + remainder;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

6(10) =


110(2)


4. Convert to the binary (base 2) the fractional part: 0.800 000 190 734 863 287.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.800 000 190 734 863 287 × 2 = 1 + 0.600 000 381 469 726 574;
  • 2) 0.600 000 381 469 726 574 × 2 = 1 + 0.200 000 762 939 453 148;
  • 3) 0.200 000 762 939 453 148 × 2 = 0 + 0.400 001 525 878 906 296;
  • 4) 0.400 001 525 878 906 296 × 2 = 0 + 0.800 003 051 757 812 592;
  • 5) 0.800 003 051 757 812 592 × 2 = 1 + 0.600 006 103 515 625 184;
  • 6) 0.600 006 103 515 625 184 × 2 = 1 + 0.200 012 207 031 250 368;
  • 7) 0.200 012 207 031 250 368 × 2 = 0 + 0.400 024 414 062 500 736;
  • 8) 0.400 024 414 062 500 736 × 2 = 0 + 0.800 048 828 125 001 472;
  • 9) 0.800 048 828 125 001 472 × 2 = 1 + 0.600 097 656 250 002 944;
  • 10) 0.600 097 656 250 002 944 × 2 = 1 + 0.200 195 312 500 005 888;
  • 11) 0.200 195 312 500 005 888 × 2 = 0 + 0.400 390 625 000 011 776;
  • 12) 0.400 390 625 000 011 776 × 2 = 0 + 0.800 781 250 000 023 552;
  • 13) 0.800 781 250 000 023 552 × 2 = 1 + 0.601 562 500 000 047 104;
  • 14) 0.601 562 500 000 047 104 × 2 = 1 + 0.203 125 000 000 094 208;
  • 15) 0.203 125 000 000 094 208 × 2 = 0 + 0.406 250 000 000 188 416;
  • 16) 0.406 250 000 000 188 416 × 2 = 0 + 0.812 500 000 000 376 832;
  • 17) 0.812 500 000 000 376 832 × 2 = 1 + 0.625 000 000 000 753 664;
  • 18) 0.625 000 000 000 753 664 × 2 = 1 + 0.250 000 000 001 507 328;
  • 19) 0.250 000 000 001 507 328 × 2 = 0 + 0.500 000 000 003 014 656;
  • 20) 0.500 000 000 003 014 656 × 2 = 1 + 0.000 000 000 006 029 312;
  • 21) 0.000 000 000 006 029 312 × 2 = 0 + 0.000 000 000 012 058 624;
  • 22) 0.000 000 000 012 058 624 × 2 = 0 + 0.000 000 000 024 117 248;
  • 23) 0.000 000 000 024 117 248 × 2 = 0 + 0.000 000 000 048 234 496;
  • 24) 0.000 000 000 048 234 496 × 2 = 0 + 0.000 000 000 096 468 992;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.800 000 190 734 863 287(10) =


0.1100 1100 1100 1100 1101 0000(2)


6. Positive number before normalization:

6.800 000 190 734 863 287(10) =


110.1100 1100 1100 1100 1101 0000(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left so that only one non zero digit remains to the left of it:

6.800 000 190 734 863 287(10) =


110.1100 1100 1100 1100 1101 0000(2) =


110.1100 1100 1100 1100 1101 0000(2) × 20 =


1.1011 0011 0011 0011 0011 0100 00(2) × 22


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 1 (a negative number)


Exponent (unadjusted): 2


Mantissa (not normalized):
1.1011 0011 0011 0011 0011 0100 00


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


2 + 2(8-1) - 1 =


(2 + 127)(10) =


129(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


129(10) =


1000 0001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 101 1001 1001 1001 1001 1010 000 =


101 1001 1001 1001 1001 1010


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (8 bits) =
1000 0001


Mantissa (23 bits) =
101 1001 1001 1001 1001 1010


Number -6.800 000 190 734 863 287 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
1 - 1000 0001 - 101 1001 1001 1001 1001 1010

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 1

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

-6.800 000 190 734 863 288 = ? ... -6.800 000 190 734 863 286 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111