-0.000 000 000 742 147 676 365 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 365(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
-0.000 000 000 742 147 676 365(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:

|-0.000 000 000 742 147 676 365| = 0.000 000 000 742 147 676 365


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: 0.000 000 000 742 147 676 365.

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) 0.000 000 000 742 147 676 365 × 2 = 0 + 0.000 000 001 484 295 352 73;
  • 2) 0.000 000 001 484 295 352 73 × 2 = 0 + 0.000 000 002 968 590 705 46;
  • 3) 0.000 000 002 968 590 705 46 × 2 = 0 + 0.000 000 005 937 181 410 92;
  • 4) 0.000 000 005 937 181 410 92 × 2 = 0 + 0.000 000 011 874 362 821 84;
  • 5) 0.000 000 011 874 362 821 84 × 2 = 0 + 0.000 000 023 748 725 643 68;
  • 6) 0.000 000 023 748 725 643 68 × 2 = 0 + 0.000 000 047 497 451 287 36;
  • 7) 0.000 000 047 497 451 287 36 × 2 = 0 + 0.000 000 094 994 902 574 72;
  • 8) 0.000 000 094 994 902 574 72 × 2 = 0 + 0.000 000 189 989 805 149 44;
  • 9) 0.000 000 189 989 805 149 44 × 2 = 0 + 0.000 000 379 979 610 298 88;
  • 10) 0.000 000 379 979 610 298 88 × 2 = 0 + 0.000 000 759 959 220 597 76;
  • 11) 0.000 000 759 959 220 597 76 × 2 = 0 + 0.000 001 519 918 441 195 52;
  • 12) 0.000 001 519 918 441 195 52 × 2 = 0 + 0.000 003 039 836 882 391 04;
  • 13) 0.000 003 039 836 882 391 04 × 2 = 0 + 0.000 006 079 673 764 782 08;
  • 14) 0.000 006 079 673 764 782 08 × 2 = 0 + 0.000 012 159 347 529 564 16;
  • 15) 0.000 012 159 347 529 564 16 × 2 = 0 + 0.000 024 318 695 059 128 32;
  • 16) 0.000 024 318 695 059 128 32 × 2 = 0 + 0.000 048 637 390 118 256 64;
  • 17) 0.000 048 637 390 118 256 64 × 2 = 0 + 0.000 097 274 780 236 513 28;
  • 18) 0.000 097 274 780 236 513 28 × 2 = 0 + 0.000 194 549 560 473 026 56;
  • 19) 0.000 194 549 560 473 026 56 × 2 = 0 + 0.000 389 099 120 946 053 12;
  • 20) 0.000 389 099 120 946 053 12 × 2 = 0 + 0.000 778 198 241 892 106 24;
  • 21) 0.000 778 198 241 892 106 24 × 2 = 0 + 0.001 556 396 483 784 212 48;
  • 22) 0.001 556 396 483 784 212 48 × 2 = 0 + 0.003 112 792 967 568 424 96;
  • 23) 0.003 112 792 967 568 424 96 × 2 = 0 + 0.006 225 585 935 136 849 92;
  • 24) 0.006 225 585 935 136 849 92 × 2 = 0 + 0.012 451 171 870 273 699 84;
  • 25) 0.012 451 171 870 273 699 84 × 2 = 0 + 0.024 902 343 740 547 399 68;
  • 26) 0.024 902 343 740 547 399 68 × 2 = 0 + 0.049 804 687 481 094 799 36;
  • 27) 0.049 804 687 481 094 799 36 × 2 = 0 + 0.099 609 374 962 189 598 72;
  • 28) 0.099 609 374 962 189 598 72 × 2 = 0 + 0.199 218 749 924 379 197 44;
  • 29) 0.199 218 749 924 379 197 44 × 2 = 0 + 0.398 437 499 848 758 394 88;
  • 30) 0.398 437 499 848 758 394 88 × 2 = 0 + 0.796 874 999 697 516 789 76;
  • 31) 0.796 874 999 697 516 789 76 × 2 = 1 + 0.593 749 999 395 033 579 52;
  • 32) 0.593 749 999 395 033 579 52 × 2 = 1 + 0.187 499 998 790 067 159 04;
  • 33) 0.187 499 998 790 067 159 04 × 2 = 0 + 0.374 999 997 580 134 318 08;
  • 34) 0.374 999 997 580 134 318 08 × 2 = 0 + 0.749 999 995 160 268 636 16;
  • 35) 0.749 999 995 160 268 636 16 × 2 = 1 + 0.499 999 990 320 537 272 32;
  • 36) 0.499 999 990 320 537 272 32 × 2 = 0 + 0.999 999 980 641 074 544 64;
  • 37) 0.999 999 980 641 074 544 64 × 2 = 1 + 0.999 999 961 282 149 089 28;
  • 38) 0.999 999 961 282 149 089 28 × 2 = 1 + 0.999 999 922 564 298 178 56;
  • 39) 0.999 999 922 564 298 178 56 × 2 = 1 + 0.999 999 845 128 596 357 12;
  • 40) 0.999 999 845 128 596 357 12 × 2 = 1 + 0.999 999 690 257 192 714 24;
  • 41) 0.999 999 690 257 192 714 24 × 2 = 1 + 0.999 999 380 514 385 428 48;
  • 42) 0.999 999 380 514 385 428 48 × 2 = 1 + 0.999 998 761 028 770 856 96;
  • 43) 0.999 998 761 028 770 856 96 × 2 = 1 + 0.999 997 522 057 541 713 92;
  • 44) 0.999 997 522 057 541 713 92 × 2 = 1 + 0.999 995 044 115 083 427 84;
  • 45) 0.999 995 044 115 083 427 84 × 2 = 1 + 0.999 990 088 230 166 855 68;
  • 46) 0.999 990 088 230 166 855 68 × 2 = 1 + 0.999 980 176 460 333 711 36;
  • 47) 0.999 980 176 460 333 711 36 × 2 = 1 + 0.999 960 352 920 667 422 72;
  • 48) 0.999 960 352 920 667 422 72 × 2 = 1 + 0.999 920 705 841 334 845 44;
  • 49) 0.999 920 705 841 334 845 44 × 2 = 1 + 0.999 841 411 682 669 690 88;
  • 50) 0.999 841 411 682 669 690 88 × 2 = 1 + 0.999 682 823 365 339 381 76;
  • 51) 0.999 682 823 365 339 381 76 × 2 = 1 + 0.999 365 646 730 678 763 52;
  • 52) 0.999 365 646 730 678 763 52 × 2 = 1 + 0.998 731 293 461 357 527 04;
  • 53) 0.998 731 293 461 357 527 04 × 2 = 1 + 0.997 462 586 922 715 054 08;
  • 54) 0.997 462 586 922 715 054 08 × 2 = 1 + 0.994 925 173 845 430 108 16;

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:


0.000 000 000 742 147 676 365(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 365(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 365(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 365 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 438 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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