-0.000 000 000 747 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 747 9(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
-0.000 000 000 747 9(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:

|-0.000 000 000 747 9| = 0.000 000 000 747 9


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 747 9.

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) 0.000 000 000 747 9 × 2 = 0 + 0.000 000 001 495 8;
  • 2) 0.000 000 001 495 8 × 2 = 0 + 0.000 000 002 991 6;
  • 3) 0.000 000 002 991 6 × 2 = 0 + 0.000 000 005 983 2;
  • 4) 0.000 000 005 983 2 × 2 = 0 + 0.000 000 011 966 4;
  • 5) 0.000 000 011 966 4 × 2 = 0 + 0.000 000 023 932 8;
  • 6) 0.000 000 023 932 8 × 2 = 0 + 0.000 000 047 865 6;
  • 7) 0.000 000 047 865 6 × 2 = 0 + 0.000 000 095 731 2;
  • 8) 0.000 000 095 731 2 × 2 = 0 + 0.000 000 191 462 4;
  • 9) 0.000 000 191 462 4 × 2 = 0 + 0.000 000 382 924 8;
  • 10) 0.000 000 382 924 8 × 2 = 0 + 0.000 000 765 849 6;
  • 11) 0.000 000 765 849 6 × 2 = 0 + 0.000 001 531 699 2;
  • 12) 0.000 001 531 699 2 × 2 = 0 + 0.000 003 063 398 4;
  • 13) 0.000 003 063 398 4 × 2 = 0 + 0.000 006 126 796 8;
  • 14) 0.000 006 126 796 8 × 2 = 0 + 0.000 012 253 593 6;
  • 15) 0.000 012 253 593 6 × 2 = 0 + 0.000 024 507 187 2;
  • 16) 0.000 024 507 187 2 × 2 = 0 + 0.000 049 014 374 4;
  • 17) 0.000 049 014 374 4 × 2 = 0 + 0.000 098 028 748 8;
  • 18) 0.000 098 028 748 8 × 2 = 0 + 0.000 196 057 497 6;
  • 19) 0.000 196 057 497 6 × 2 = 0 + 0.000 392 114 995 2;
  • 20) 0.000 392 114 995 2 × 2 = 0 + 0.000 784 229 990 4;
  • 21) 0.000 784 229 990 4 × 2 = 0 + 0.001 568 459 980 8;
  • 22) 0.001 568 459 980 8 × 2 = 0 + 0.003 136 919 961 6;
  • 23) 0.003 136 919 961 6 × 2 = 0 + 0.006 273 839 923 2;
  • 24) 0.006 273 839 923 2 × 2 = 0 + 0.012 547 679 846 4;
  • 25) 0.012 547 679 846 4 × 2 = 0 + 0.025 095 359 692 8;
  • 26) 0.025 095 359 692 8 × 2 = 0 + 0.050 190 719 385 6;
  • 27) 0.050 190 719 385 6 × 2 = 0 + 0.100 381 438 771 2;
  • 28) 0.100 381 438 771 2 × 2 = 0 + 0.200 762 877 542 4;
  • 29) 0.200 762 877 542 4 × 2 = 0 + 0.401 525 755 084 8;
  • 30) 0.401 525 755 084 8 × 2 = 0 + 0.803 051 510 169 6;
  • 31) 0.803 051 510 169 6 × 2 = 1 + 0.606 103 020 339 2;
  • 32) 0.606 103 020 339 2 × 2 = 1 + 0.212 206 040 678 4;
  • 33) 0.212 206 040 678 4 × 2 = 0 + 0.424 412 081 356 8;
  • 34) 0.424 412 081 356 8 × 2 = 0 + 0.848 824 162 713 6;
  • 35) 0.848 824 162 713 6 × 2 = 1 + 0.697 648 325 427 2;
  • 36) 0.697 648 325 427 2 × 2 = 1 + 0.395 296 650 854 4;
  • 37) 0.395 296 650 854 4 × 2 = 0 + 0.790 593 301 708 8;
  • 38) 0.790 593 301 708 8 × 2 = 1 + 0.581 186 603 417 6;
  • 39) 0.581 186 603 417 6 × 2 = 1 + 0.162 373 206 835 2;
  • 40) 0.162 373 206 835 2 × 2 = 0 + 0.324 746 413 670 4;
  • 41) 0.324 746 413 670 4 × 2 = 0 + 0.649 492 827 340 8;
  • 42) 0.649 492 827 340 8 × 2 = 1 + 0.298 985 654 681 6;
  • 43) 0.298 985 654 681 6 × 2 = 0 + 0.597 971 309 363 2;
  • 44) 0.597 971 309 363 2 × 2 = 1 + 0.195 942 618 726 4;
  • 45) 0.195 942 618 726 4 × 2 = 0 + 0.391 885 237 452 8;
  • 46) 0.391 885 237 452 8 × 2 = 0 + 0.783 770 474 905 6;
  • 47) 0.783 770 474 905 6 × 2 = 1 + 0.567 540 949 811 2;
  • 48) 0.567 540 949 811 2 × 2 = 1 + 0.135 081 899 622 4;
  • 49) 0.135 081 899 622 4 × 2 = 0 + 0.270 163 799 244 8;
  • 50) 0.270 163 799 244 8 × 2 = 0 + 0.540 327 598 489 6;
  • 51) 0.540 327 598 489 6 × 2 = 1 + 0.080 655 196 979 2;
  • 52) 0.080 655 196 979 2 × 2 = 0 + 0.161 310 393 958 4;
  • 53) 0.161 310 393 958 4 × 2 = 0 + 0.322 620 787 916 8;
  • 54) 0.322 620 787 916 8 × 2 = 0 + 0.645 241 575 833 6;

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:


0.000 000 000 747 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 0101 0011 0010 00(2)

6. Positive number before normalization:

0.000 000 000 747 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 0101 0011 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 747 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 0101 0011 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 0101 0011 0010 00(2) × 20 =

1.1001 1011 0010 1001 1001 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1011 0010 1001 1001 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1101 1001 0100 1100 1000 =

100 1101 1001 0100 1100 1000


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) =
0110 0000

Mantissa (23 bits) =
100 1101 1001 0100 1100 1000


Decimal number -0.000 000 000 747 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 1001 0100 1100 1000

Share

» Convert decimal -0.000 000 000 741 9 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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