-500.311 000 011 111 110 100 010 015 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -500.311 000 011 111 110 100 010 015 5(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-500.311 000 011 111 110 100 010 015 5(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-500.311 000 011 111 110 100 010 015 5| = 500.311 000 011 111 110 100 010 015 5


2. First, convert to binary (in base 2) the integer part: 500.
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;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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.

500(10) =

1 1111 0100(2)


4. Convert to binary (base 2) the fractional part: 0.311 000 011 111 110 100 010 015 5.

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) 0.311 000 011 111 110 100 010 015 5 × 2 = 0 + 0.622 000 022 222 220 200 020 031;
  • 2) 0.622 000 022 222 220 200 020 031 × 2 = 1 + 0.244 000 044 444 440 400 040 062;
  • 3) 0.244 000 044 444 440 400 040 062 × 2 = 0 + 0.488 000 088 888 880 800 080 124;
  • 4) 0.488 000 088 888 880 800 080 124 × 2 = 0 + 0.976 000 177 777 761 600 160 248;
  • 5) 0.976 000 177 777 761 600 160 248 × 2 = 1 + 0.952 000 355 555 523 200 320 496;
  • 6) 0.952 000 355 555 523 200 320 496 × 2 = 1 + 0.904 000 711 111 046 400 640 992;
  • 7) 0.904 000 711 111 046 400 640 992 × 2 = 1 + 0.808 001 422 222 092 801 281 984;
  • 8) 0.808 001 422 222 092 801 281 984 × 2 = 1 + 0.616 002 844 444 185 602 563 968;
  • 9) 0.616 002 844 444 185 602 563 968 × 2 = 1 + 0.232 005 688 888 371 205 127 936;
  • 10) 0.232 005 688 888 371 205 127 936 × 2 = 0 + 0.464 011 377 776 742 410 255 872;
  • 11) 0.464 011 377 776 742 410 255 872 × 2 = 0 + 0.928 022 755 553 484 820 511 744;
  • 12) 0.928 022 755 553 484 820 511 744 × 2 = 1 + 0.856 045 511 106 969 641 023 488;
  • 13) 0.856 045 511 106 969 641 023 488 × 2 = 1 + 0.712 091 022 213 939 282 046 976;
  • 14) 0.712 091 022 213 939 282 046 976 × 2 = 1 + 0.424 182 044 427 878 564 093 952;
  • 15) 0.424 182 044 427 878 564 093 952 × 2 = 0 + 0.848 364 088 855 757 128 187 904;
  • 16) 0.848 364 088 855 757 128 187 904 × 2 = 1 + 0.696 728 177 711 514 256 375 808;
  • 17) 0.696 728 177 711 514 256 375 808 × 2 = 1 + 0.393 456 355 423 028 512 751 616;
  • 18) 0.393 456 355 423 028 512 751 616 × 2 = 0 + 0.786 912 710 846 057 025 503 232;
  • 19) 0.786 912 710 846 057 025 503 232 × 2 = 1 + 0.573 825 421 692 114 051 006 464;
  • 20) 0.573 825 421 692 114 051 006 464 × 2 = 1 + 0.147 650 843 384 228 102 012 928;
  • 21) 0.147 650 843 384 228 102 012 928 × 2 = 0 + 0.295 301 686 768 456 204 025 856;
  • 22) 0.295 301 686 768 456 204 025 856 × 2 = 0 + 0.590 603 373 536 912 408 051 712;
  • 23) 0.590 603 373 536 912 408 051 712 × 2 = 1 + 0.181 206 747 073 824 816 103 424;
  • 24) 0.181 206 747 073 824 816 103 424 × 2 = 0 + 0.362 413 494 147 649 632 206 848;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.311 000 011 111 110 100 010 015 5(10) =

0.0100 1111 1001 1101 1011 0010(2)

6. Positive number before normalization:

500.311 000 011 111 110 100 010 015 5(10) =

1 1111 0100.0100 1111 1001 1101 1011 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left, so that only one non zero digit remains to the left of it:


500.311 000 011 111 110 100 010 015 5(10) =

1 1111 0100.0100 1111 1001 1101 1011 0010(2) =

1 1111 0100.0100 1111 1001 1101 1011 0010(2) × 20 =

1.1111 0100 0100 1111 1001 1101 1011 0010(2) × 28


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): 8

Mantissa (not normalized):
1.1111 0100 0100 1111 1001 1101 1011 0010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

8 + 2(8-1) - 1 =

(8 + 127)(10) =

135(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;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 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) =

135(10) =

1000 0111(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. 111 1010 0010 0111 1100 1110 1 1011 0010 =

111 1010 0010 0111 1100 1110


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 0111

Mantissa (23 bits) =
111 1010 0010 0111 1100 1110


Decimal number -500.311 000 011 111 110 100 010 015 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0111 - 111 1010 0010 0111 1100 1110

Share

» Convert decimal -500.311 000 011 111 110 100 010 009 2 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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