-0.000 000 000 742 147 676 795 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 795(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
-0.000 000 000 742 147 676 795(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:

|-0.000 000 000 742 147 676 795| = 0.000 000 000 742 147 676 795


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: 0.000 000 000 742 147 676 795.

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) 0.000 000 000 742 147 676 795 × 2 = 0 + 0.000 000 001 484 295 353 59;
  • 2) 0.000 000 001 484 295 353 59 × 2 = 0 + 0.000 000 002 968 590 707 18;
  • 3) 0.000 000 002 968 590 707 18 × 2 = 0 + 0.000 000 005 937 181 414 36;
  • 4) 0.000 000 005 937 181 414 36 × 2 = 0 + 0.000 000 011 874 362 828 72;
  • 5) 0.000 000 011 874 362 828 72 × 2 = 0 + 0.000 000 023 748 725 657 44;
  • 6) 0.000 000 023 748 725 657 44 × 2 = 0 + 0.000 000 047 497 451 314 88;
  • 7) 0.000 000 047 497 451 314 88 × 2 = 0 + 0.000 000 094 994 902 629 76;
  • 8) 0.000 000 094 994 902 629 76 × 2 = 0 + 0.000 000 189 989 805 259 52;
  • 9) 0.000 000 189 989 805 259 52 × 2 = 0 + 0.000 000 379 979 610 519 04;
  • 10) 0.000 000 379 979 610 519 04 × 2 = 0 + 0.000 000 759 959 221 038 08;
  • 11) 0.000 000 759 959 221 038 08 × 2 = 0 + 0.000 001 519 918 442 076 16;
  • 12) 0.000 001 519 918 442 076 16 × 2 = 0 + 0.000 003 039 836 884 152 32;
  • 13) 0.000 003 039 836 884 152 32 × 2 = 0 + 0.000 006 079 673 768 304 64;
  • 14) 0.000 006 079 673 768 304 64 × 2 = 0 + 0.000 012 159 347 536 609 28;
  • 15) 0.000 012 159 347 536 609 28 × 2 = 0 + 0.000 024 318 695 073 218 56;
  • 16) 0.000 024 318 695 073 218 56 × 2 = 0 + 0.000 048 637 390 146 437 12;
  • 17) 0.000 048 637 390 146 437 12 × 2 = 0 + 0.000 097 274 780 292 874 24;
  • 18) 0.000 097 274 780 292 874 24 × 2 = 0 + 0.000 194 549 560 585 748 48;
  • 19) 0.000 194 549 560 585 748 48 × 2 = 0 + 0.000 389 099 121 171 496 96;
  • 20) 0.000 389 099 121 171 496 96 × 2 = 0 + 0.000 778 198 242 342 993 92;
  • 21) 0.000 778 198 242 342 993 92 × 2 = 0 + 0.001 556 396 484 685 987 84;
  • 22) 0.001 556 396 484 685 987 84 × 2 = 0 + 0.003 112 792 969 371 975 68;
  • 23) 0.003 112 792 969 371 975 68 × 2 = 0 + 0.006 225 585 938 743 951 36;
  • 24) 0.006 225 585 938 743 951 36 × 2 = 0 + 0.012 451 171 877 487 902 72;
  • 25) 0.012 451 171 877 487 902 72 × 2 = 0 + 0.024 902 343 754 975 805 44;
  • 26) 0.024 902 343 754 975 805 44 × 2 = 0 + 0.049 804 687 509 951 610 88;
  • 27) 0.049 804 687 509 951 610 88 × 2 = 0 + 0.099 609 375 019 903 221 76;
  • 28) 0.099 609 375 019 903 221 76 × 2 = 0 + 0.199 218 750 039 806 443 52;
  • 29) 0.199 218 750 039 806 443 52 × 2 = 0 + 0.398 437 500 079 612 887 04;
  • 30) 0.398 437 500 079 612 887 04 × 2 = 0 + 0.796 875 000 159 225 774 08;
  • 31) 0.796 875 000 159 225 774 08 × 2 = 1 + 0.593 750 000 318 451 548 16;
  • 32) 0.593 750 000 318 451 548 16 × 2 = 1 + 0.187 500 000 636 903 096 32;
  • 33) 0.187 500 000 636 903 096 32 × 2 = 0 + 0.375 000 001 273 806 192 64;
  • 34) 0.375 000 001 273 806 192 64 × 2 = 0 + 0.750 000 002 547 612 385 28;
  • 35) 0.750 000 002 547 612 385 28 × 2 = 1 + 0.500 000 005 095 224 770 56;
  • 36) 0.500 000 005 095 224 770 56 × 2 = 1 + 0.000 000 010 190 449 541 12;
  • 37) 0.000 000 010 190 449 541 12 × 2 = 0 + 0.000 000 020 380 899 082 24;
  • 38) 0.000 000 020 380 899 082 24 × 2 = 0 + 0.000 000 040 761 798 164 48;
  • 39) 0.000 000 040 761 798 164 48 × 2 = 0 + 0.000 000 081 523 596 328 96;
  • 40) 0.000 000 081 523 596 328 96 × 2 = 0 + 0.000 000 163 047 192 657 92;
  • 41) 0.000 000 163 047 192 657 92 × 2 = 0 + 0.000 000 326 094 385 315 84;
  • 42) 0.000 000 326 094 385 315 84 × 2 = 0 + 0.000 000 652 188 770 631 68;
  • 43) 0.000 000 652 188 770 631 68 × 2 = 0 + 0.000 001 304 377 541 263 36;
  • 44) 0.000 001 304 377 541 263 36 × 2 = 0 + 0.000 002 608 755 082 526 72;
  • 45) 0.000 002 608 755 082 526 72 × 2 = 0 + 0.000 005 217 510 165 053 44;
  • 46) 0.000 005 217 510 165 053 44 × 2 = 0 + 0.000 010 435 020 330 106 88;
  • 47) 0.000 010 435 020 330 106 88 × 2 = 0 + 0.000 020 870 040 660 213 76;
  • 48) 0.000 020 870 040 660 213 76 × 2 = 0 + 0.000 041 740 081 320 427 52;
  • 49) 0.000 041 740 081 320 427 52 × 2 = 0 + 0.000 083 480 162 640 855 04;
  • 50) 0.000 083 480 162 640 855 04 × 2 = 0 + 0.000 166 960 325 281 710 08;
  • 51) 0.000 166 960 325 281 710 08 × 2 = 0 + 0.000 333 920 650 563 420 16;
  • 52) 0.000 333 920 650 563 420 16 × 2 = 0 + 0.000 667 841 301 126 840 32;
  • 53) 0.000 667 841 301 126 840 32 × 2 = 0 + 0.001 335 682 602 253 680 64;
  • 54) 0.001 335 682 602 253 680 64 × 2 = 0 + 0.002 671 365 204 507 361 28;

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:


0.000 000 000 742 147 676 795(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 795(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:


0.000 000 000 742 147 676 795(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 -0.000 000 000 742 147 676 795 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