-0.000 000 000 742 147 677 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 93(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
-0.000 000 000 742 147 677 93(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:

|-0.000 000 000 742 147 677 93| = 0.000 000 000 742 147 677 93


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: 0.000 000 000 742 147 677 93.

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) 0.000 000 000 742 147 677 93 × 2 = 0 + 0.000 000 001 484 295 355 86;
  • 2) 0.000 000 001 484 295 355 86 × 2 = 0 + 0.000 000 002 968 590 711 72;
  • 3) 0.000 000 002 968 590 711 72 × 2 = 0 + 0.000 000 005 937 181 423 44;
  • 4) 0.000 000 005 937 181 423 44 × 2 = 0 + 0.000 000 011 874 362 846 88;
  • 5) 0.000 000 011 874 362 846 88 × 2 = 0 + 0.000 000 023 748 725 693 76;
  • 6) 0.000 000 023 748 725 693 76 × 2 = 0 + 0.000 000 047 497 451 387 52;
  • 7) 0.000 000 047 497 451 387 52 × 2 = 0 + 0.000 000 094 994 902 775 04;
  • 8) 0.000 000 094 994 902 775 04 × 2 = 0 + 0.000 000 189 989 805 550 08;
  • 9) 0.000 000 189 989 805 550 08 × 2 = 0 + 0.000 000 379 979 611 100 16;
  • 10) 0.000 000 379 979 611 100 16 × 2 = 0 + 0.000 000 759 959 222 200 32;
  • 11) 0.000 000 759 959 222 200 32 × 2 = 0 + 0.000 001 519 918 444 400 64;
  • 12) 0.000 001 519 918 444 400 64 × 2 = 0 + 0.000 003 039 836 888 801 28;
  • 13) 0.000 003 039 836 888 801 28 × 2 = 0 + 0.000 006 079 673 777 602 56;
  • 14) 0.000 006 079 673 777 602 56 × 2 = 0 + 0.000 012 159 347 555 205 12;
  • 15) 0.000 012 159 347 555 205 12 × 2 = 0 + 0.000 024 318 695 110 410 24;
  • 16) 0.000 024 318 695 110 410 24 × 2 = 0 + 0.000 048 637 390 220 820 48;
  • 17) 0.000 048 637 390 220 820 48 × 2 = 0 + 0.000 097 274 780 441 640 96;
  • 18) 0.000 097 274 780 441 640 96 × 2 = 0 + 0.000 194 549 560 883 281 92;
  • 19) 0.000 194 549 560 883 281 92 × 2 = 0 + 0.000 389 099 121 766 563 84;
  • 20) 0.000 389 099 121 766 563 84 × 2 = 0 + 0.000 778 198 243 533 127 68;
  • 21) 0.000 778 198 243 533 127 68 × 2 = 0 + 0.001 556 396 487 066 255 36;
  • 22) 0.001 556 396 487 066 255 36 × 2 = 0 + 0.003 112 792 974 132 510 72;
  • 23) 0.003 112 792 974 132 510 72 × 2 = 0 + 0.006 225 585 948 265 021 44;
  • 24) 0.006 225 585 948 265 021 44 × 2 = 0 + 0.012 451 171 896 530 042 88;
  • 25) 0.012 451 171 896 530 042 88 × 2 = 0 + 0.024 902 343 793 060 085 76;
  • 26) 0.024 902 343 793 060 085 76 × 2 = 0 + 0.049 804 687 586 120 171 52;
  • 27) 0.049 804 687 586 120 171 52 × 2 = 0 + 0.099 609 375 172 240 343 04;
  • 28) 0.099 609 375 172 240 343 04 × 2 = 0 + 0.199 218 750 344 480 686 08;
  • 29) 0.199 218 750 344 480 686 08 × 2 = 0 + 0.398 437 500 688 961 372 16;
  • 30) 0.398 437 500 688 961 372 16 × 2 = 0 + 0.796 875 001 377 922 744 32;
  • 31) 0.796 875 001 377 922 744 32 × 2 = 1 + 0.593 750 002 755 845 488 64;
  • 32) 0.593 750 002 755 845 488 64 × 2 = 1 + 0.187 500 005 511 690 977 28;
  • 33) 0.187 500 005 511 690 977 28 × 2 = 0 + 0.375 000 011 023 381 954 56;
  • 34) 0.375 000 011 023 381 954 56 × 2 = 0 + 0.750 000 022 046 763 909 12;
  • 35) 0.750 000 022 046 763 909 12 × 2 = 1 + 0.500 000 044 093 527 818 24;
  • 36) 0.500 000 044 093 527 818 24 × 2 = 1 + 0.000 000 088 187 055 636 48;
  • 37) 0.000 000 088 187 055 636 48 × 2 = 0 + 0.000 000 176 374 111 272 96;
  • 38) 0.000 000 176 374 111 272 96 × 2 = 0 + 0.000 000 352 748 222 545 92;
  • 39) 0.000 000 352 748 222 545 92 × 2 = 0 + 0.000 000 705 496 445 091 84;
  • 40) 0.000 000 705 496 445 091 84 × 2 = 0 + 0.000 001 410 992 890 183 68;
  • 41) 0.000 001 410 992 890 183 68 × 2 = 0 + 0.000 002 821 985 780 367 36;
  • 42) 0.000 002 821 985 780 367 36 × 2 = 0 + 0.000 005 643 971 560 734 72;
  • 43) 0.000 005 643 971 560 734 72 × 2 = 0 + 0.000 011 287 943 121 469 44;
  • 44) 0.000 011 287 943 121 469 44 × 2 = 0 + 0.000 022 575 886 242 938 88;
  • 45) 0.000 022 575 886 242 938 88 × 2 = 0 + 0.000 045 151 772 485 877 76;
  • 46) 0.000 045 151 772 485 877 76 × 2 = 0 + 0.000 090 303 544 971 755 52;
  • 47) 0.000 090 303 544 971 755 52 × 2 = 0 + 0.000 180 607 089 943 511 04;
  • 48) 0.000 180 607 089 943 511 04 × 2 = 0 + 0.000 361 214 179 887 022 08;
  • 49) 0.000 361 214 179 887 022 08 × 2 = 0 + 0.000 722 428 359 774 044 16;
  • 50) 0.000 722 428 359 774 044 16 × 2 = 0 + 0.001 444 856 719 548 088 32;
  • 51) 0.001 444 856 719 548 088 32 × 2 = 0 + 0.002 889 713 439 096 176 64;
  • 52) 0.002 889 713 439 096 176 64 × 2 = 0 + 0.005 779 426 878 192 353 28;
  • 53) 0.005 779 426 878 192 353 28 × 2 = 0 + 0.011 558 853 756 384 706 56;
  • 54) 0.011 558 853 756 384 706 56 × 2 = 0 + 0.023 117 707 512 769 413 12;

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:


0.000 000 000 742 147 677 93(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 93(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:


0.000 000 000 742 147 677 93(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 -0.000 000 000 742 147 677 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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