-0.000 000 000 742 147 676 646 76 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 76(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
-0.000 000 000 742 147 676 646 76(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:

|-0.000 000 000 742 147 676 646 76| = 0.000 000 000 742 147 676 646 76


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: 0.000 000 000 742 147 676 646 76.

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) 0.000 000 000 742 147 676 646 76 × 2 = 0 + 0.000 000 001 484 295 353 293 52;
  • 2) 0.000 000 001 484 295 353 293 52 × 2 = 0 + 0.000 000 002 968 590 706 587 04;
  • 3) 0.000 000 002 968 590 706 587 04 × 2 = 0 + 0.000 000 005 937 181 413 174 08;
  • 4) 0.000 000 005 937 181 413 174 08 × 2 = 0 + 0.000 000 011 874 362 826 348 16;
  • 5) 0.000 000 011 874 362 826 348 16 × 2 = 0 + 0.000 000 023 748 725 652 696 32;
  • 6) 0.000 000 023 748 725 652 696 32 × 2 = 0 + 0.000 000 047 497 451 305 392 64;
  • 7) 0.000 000 047 497 451 305 392 64 × 2 = 0 + 0.000 000 094 994 902 610 785 28;
  • 8) 0.000 000 094 994 902 610 785 28 × 2 = 0 + 0.000 000 189 989 805 221 570 56;
  • 9) 0.000 000 189 989 805 221 570 56 × 2 = 0 + 0.000 000 379 979 610 443 141 12;
  • 10) 0.000 000 379 979 610 443 141 12 × 2 = 0 + 0.000 000 759 959 220 886 282 24;
  • 11) 0.000 000 759 959 220 886 282 24 × 2 = 0 + 0.000 001 519 918 441 772 564 48;
  • 12) 0.000 001 519 918 441 772 564 48 × 2 = 0 + 0.000 003 039 836 883 545 128 96;
  • 13) 0.000 003 039 836 883 545 128 96 × 2 = 0 + 0.000 006 079 673 767 090 257 92;
  • 14) 0.000 006 079 673 767 090 257 92 × 2 = 0 + 0.000 012 159 347 534 180 515 84;
  • 15) 0.000 012 159 347 534 180 515 84 × 2 = 0 + 0.000 024 318 695 068 361 031 68;
  • 16) 0.000 024 318 695 068 361 031 68 × 2 = 0 + 0.000 048 637 390 136 722 063 36;
  • 17) 0.000 048 637 390 136 722 063 36 × 2 = 0 + 0.000 097 274 780 273 444 126 72;
  • 18) 0.000 097 274 780 273 444 126 72 × 2 = 0 + 0.000 194 549 560 546 888 253 44;
  • 19) 0.000 194 549 560 546 888 253 44 × 2 = 0 + 0.000 389 099 121 093 776 506 88;
  • 20) 0.000 389 099 121 093 776 506 88 × 2 = 0 + 0.000 778 198 242 187 553 013 76;
  • 21) 0.000 778 198 242 187 553 013 76 × 2 = 0 + 0.001 556 396 484 375 106 027 52;
  • 22) 0.001 556 396 484 375 106 027 52 × 2 = 0 + 0.003 112 792 968 750 212 055 04;
  • 23) 0.003 112 792 968 750 212 055 04 × 2 = 0 + 0.006 225 585 937 500 424 110 08;
  • 24) 0.006 225 585 937 500 424 110 08 × 2 = 0 + 0.012 451 171 875 000 848 220 16;
  • 25) 0.012 451 171 875 000 848 220 16 × 2 = 0 + 0.024 902 343 750 001 696 440 32;
  • 26) 0.024 902 343 750 001 696 440 32 × 2 = 0 + 0.049 804 687 500 003 392 880 64;
  • 27) 0.049 804 687 500 003 392 880 64 × 2 = 0 + 0.099 609 375 000 006 785 761 28;
  • 28) 0.099 609 375 000 006 785 761 28 × 2 = 0 + 0.199 218 750 000 013 571 522 56;
  • 29) 0.199 218 750 000 013 571 522 56 × 2 = 0 + 0.398 437 500 000 027 143 045 12;
  • 30) 0.398 437 500 000 027 143 045 12 × 2 = 0 + 0.796 875 000 000 054 286 090 24;
  • 31) 0.796 875 000 000 054 286 090 24 × 2 = 1 + 0.593 750 000 000 108 572 180 48;
  • 32) 0.593 750 000 000 108 572 180 48 × 2 = 1 + 0.187 500 000 000 217 144 360 96;
  • 33) 0.187 500 000 000 217 144 360 96 × 2 = 0 + 0.375 000 000 000 434 288 721 92;
  • 34) 0.375 000 000 000 434 288 721 92 × 2 = 0 + 0.750 000 000 000 868 577 443 84;
  • 35) 0.750 000 000 000 868 577 443 84 × 2 = 1 + 0.500 000 000 001 737 154 887 68;
  • 36) 0.500 000 000 001 737 154 887 68 × 2 = 1 + 0.000 000 000 003 474 309 775 36;
  • 37) 0.000 000 000 003 474 309 775 36 × 2 = 0 + 0.000 000 000 006 948 619 550 72;
  • 38) 0.000 000 000 006 948 619 550 72 × 2 = 0 + 0.000 000 000 013 897 239 101 44;
  • 39) 0.000 000 000 013 897 239 101 44 × 2 = 0 + 0.000 000 000 027 794 478 202 88;
  • 40) 0.000 000 000 027 794 478 202 88 × 2 = 0 + 0.000 000 000 055 588 956 405 76;
  • 41) 0.000 000 000 055 588 956 405 76 × 2 = 0 + 0.000 000 000 111 177 912 811 52;
  • 42) 0.000 000 000 111 177 912 811 52 × 2 = 0 + 0.000 000 000 222 355 825 623 04;
  • 43) 0.000 000 000 222 355 825 623 04 × 2 = 0 + 0.000 000 000 444 711 651 246 08;
  • 44) 0.000 000 000 444 711 651 246 08 × 2 = 0 + 0.000 000 000 889 423 302 492 16;
  • 45) 0.000 000 000 889 423 302 492 16 × 2 = 0 + 0.000 000 001 778 846 604 984 32;
  • 46) 0.000 000 001 778 846 604 984 32 × 2 = 0 + 0.000 000 003 557 693 209 968 64;
  • 47) 0.000 000 003 557 693 209 968 64 × 2 = 0 + 0.000 000 007 115 386 419 937 28;
  • 48) 0.000 000 007 115 386 419 937 28 × 2 = 0 + 0.000 000 014 230 772 839 874 56;
  • 49) 0.000 000 014 230 772 839 874 56 × 2 = 0 + 0.000 000 028 461 545 679 749 12;
  • 50) 0.000 000 028 461 545 679 749 12 × 2 = 0 + 0.000 000 056 923 091 359 498 24;
  • 51) 0.000 000 056 923 091 359 498 24 × 2 = 0 + 0.000 000 113 846 182 718 996 48;
  • 52) 0.000 000 113 846 182 718 996 48 × 2 = 0 + 0.000 000 227 692 365 437 992 96;
  • 53) 0.000 000 227 692 365 437 992 96 × 2 = 0 + 0.000 000 455 384 730 875 985 92;
  • 54) 0.000 000 455 384 730 875 985 92 × 2 = 0 + 0.000 000 910 769 461 751 971 84;

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:


0.000 000 000 742 147 676 646 76(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 76(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:


0.000 000 000 742 147 676 646 76(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 -0.000 000 000 742 147 676 646 76 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 645 96 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: 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