-0.000 000 000 742 147 691 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 691 1(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
-0.000 000 000 742 147 691 1(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:

|-0.000 000 000 742 147 691 1| = 0.000 000 000 742 147 691 1


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: 0.000 000 000 742 147 691 1.

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) 0.000 000 000 742 147 691 1 × 2 = 0 + 0.000 000 001 484 295 382 2;
  • 2) 0.000 000 001 484 295 382 2 × 2 = 0 + 0.000 000 002 968 590 764 4;
  • 3) 0.000 000 002 968 590 764 4 × 2 = 0 + 0.000 000 005 937 181 528 8;
  • 4) 0.000 000 005 937 181 528 8 × 2 = 0 + 0.000 000 011 874 363 057 6;
  • 5) 0.000 000 011 874 363 057 6 × 2 = 0 + 0.000 000 023 748 726 115 2;
  • 6) 0.000 000 023 748 726 115 2 × 2 = 0 + 0.000 000 047 497 452 230 4;
  • 7) 0.000 000 047 497 452 230 4 × 2 = 0 + 0.000 000 094 994 904 460 8;
  • 8) 0.000 000 094 994 904 460 8 × 2 = 0 + 0.000 000 189 989 808 921 6;
  • 9) 0.000 000 189 989 808 921 6 × 2 = 0 + 0.000 000 379 979 617 843 2;
  • 10) 0.000 000 379 979 617 843 2 × 2 = 0 + 0.000 000 759 959 235 686 4;
  • 11) 0.000 000 759 959 235 686 4 × 2 = 0 + 0.000 001 519 918 471 372 8;
  • 12) 0.000 001 519 918 471 372 8 × 2 = 0 + 0.000 003 039 836 942 745 6;
  • 13) 0.000 003 039 836 942 745 6 × 2 = 0 + 0.000 006 079 673 885 491 2;
  • 14) 0.000 006 079 673 885 491 2 × 2 = 0 + 0.000 012 159 347 770 982 4;
  • 15) 0.000 012 159 347 770 982 4 × 2 = 0 + 0.000 024 318 695 541 964 8;
  • 16) 0.000 024 318 695 541 964 8 × 2 = 0 + 0.000 048 637 391 083 929 6;
  • 17) 0.000 048 637 391 083 929 6 × 2 = 0 + 0.000 097 274 782 167 859 2;
  • 18) 0.000 097 274 782 167 859 2 × 2 = 0 + 0.000 194 549 564 335 718 4;
  • 19) 0.000 194 549 564 335 718 4 × 2 = 0 + 0.000 389 099 128 671 436 8;
  • 20) 0.000 389 099 128 671 436 8 × 2 = 0 + 0.000 778 198 257 342 873 6;
  • 21) 0.000 778 198 257 342 873 6 × 2 = 0 + 0.001 556 396 514 685 747 2;
  • 22) 0.001 556 396 514 685 747 2 × 2 = 0 + 0.003 112 793 029 371 494 4;
  • 23) 0.003 112 793 029 371 494 4 × 2 = 0 + 0.006 225 586 058 742 988 8;
  • 24) 0.006 225 586 058 742 988 8 × 2 = 0 + 0.012 451 172 117 485 977 6;
  • 25) 0.012 451 172 117 485 977 6 × 2 = 0 + 0.024 902 344 234 971 955 2;
  • 26) 0.024 902 344 234 971 955 2 × 2 = 0 + 0.049 804 688 469 943 910 4;
  • 27) 0.049 804 688 469 943 910 4 × 2 = 0 + 0.099 609 376 939 887 820 8;
  • 28) 0.099 609 376 939 887 820 8 × 2 = 0 + 0.199 218 753 879 775 641 6;
  • 29) 0.199 218 753 879 775 641 6 × 2 = 0 + 0.398 437 507 759 551 283 2;
  • 30) 0.398 437 507 759 551 283 2 × 2 = 0 + 0.796 875 015 519 102 566 4;
  • 31) 0.796 875 015 519 102 566 4 × 2 = 1 + 0.593 750 031 038 205 132 8;
  • 32) 0.593 750 031 038 205 132 8 × 2 = 1 + 0.187 500 062 076 410 265 6;
  • 33) 0.187 500 062 076 410 265 6 × 2 = 0 + 0.375 000 124 152 820 531 2;
  • 34) 0.375 000 124 152 820 531 2 × 2 = 0 + 0.750 000 248 305 641 062 4;
  • 35) 0.750 000 248 305 641 062 4 × 2 = 1 + 0.500 000 496 611 282 124 8;
  • 36) 0.500 000 496 611 282 124 8 × 2 = 1 + 0.000 000 993 222 564 249 6;
  • 37) 0.000 000 993 222 564 249 6 × 2 = 0 + 0.000 001 986 445 128 499 2;
  • 38) 0.000 001 986 445 128 499 2 × 2 = 0 + 0.000 003 972 890 256 998 4;
  • 39) 0.000 003 972 890 256 998 4 × 2 = 0 + 0.000 007 945 780 513 996 8;
  • 40) 0.000 007 945 780 513 996 8 × 2 = 0 + 0.000 015 891 561 027 993 6;
  • 41) 0.000 015 891 561 027 993 6 × 2 = 0 + 0.000 031 783 122 055 987 2;
  • 42) 0.000 031 783 122 055 987 2 × 2 = 0 + 0.000 063 566 244 111 974 4;
  • 43) 0.000 063 566 244 111 974 4 × 2 = 0 + 0.000 127 132 488 223 948 8;
  • 44) 0.000 127 132 488 223 948 8 × 2 = 0 + 0.000 254 264 976 447 897 6;
  • 45) 0.000 254 264 976 447 897 6 × 2 = 0 + 0.000 508 529 952 895 795 2;
  • 46) 0.000 508 529 952 895 795 2 × 2 = 0 + 0.001 017 059 905 791 590 4;
  • 47) 0.001 017 059 905 791 590 4 × 2 = 0 + 0.002 034 119 811 583 180 8;
  • 48) 0.002 034 119 811 583 180 8 × 2 = 0 + 0.004 068 239 623 166 361 6;
  • 49) 0.004 068 239 623 166 361 6 × 2 = 0 + 0.008 136 479 246 332 723 2;
  • 50) 0.008 136 479 246 332 723 2 × 2 = 0 + 0.016 272 958 492 665 446 4;
  • 51) 0.016 272 958 492 665 446 4 × 2 = 0 + 0.032 545 916 985 330 892 8;
  • 52) 0.032 545 916 985 330 892 8 × 2 = 0 + 0.065 091 833 970 661 785 6;
  • 53) 0.065 091 833 970 661 785 6 × 2 = 0 + 0.130 183 667 941 323 571 2;
  • 54) 0.130 183 667 941 323 571 2 × 2 = 0 + 0.260 367 335 882 647 142 4;

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:


0.000 000 000 742 147 691 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 691 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 691 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 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 1000 0000 0000 0000 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 691 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 682 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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