-0.000 000 000 742 147 677 06 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 06(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 06(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 06| = 0.000 000 000 742 147 677 06


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 06.

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 06 × 2 = 0 + 0.000 000 001 484 295 354 12;
  • 2) 0.000 000 001 484 295 354 12 × 2 = 0 + 0.000 000 002 968 590 708 24;
  • 3) 0.000 000 002 968 590 708 24 × 2 = 0 + 0.000 000 005 937 181 416 48;
  • 4) 0.000 000 005 937 181 416 48 × 2 = 0 + 0.000 000 011 874 362 832 96;
  • 5) 0.000 000 011 874 362 832 96 × 2 = 0 + 0.000 000 023 748 725 665 92;
  • 6) 0.000 000 023 748 725 665 92 × 2 = 0 + 0.000 000 047 497 451 331 84;
  • 7) 0.000 000 047 497 451 331 84 × 2 = 0 + 0.000 000 094 994 902 663 68;
  • 8) 0.000 000 094 994 902 663 68 × 2 = 0 + 0.000 000 189 989 805 327 36;
  • 9) 0.000 000 189 989 805 327 36 × 2 = 0 + 0.000 000 379 979 610 654 72;
  • 10) 0.000 000 379 979 610 654 72 × 2 = 0 + 0.000 000 759 959 221 309 44;
  • 11) 0.000 000 759 959 221 309 44 × 2 = 0 + 0.000 001 519 918 442 618 88;
  • 12) 0.000 001 519 918 442 618 88 × 2 = 0 + 0.000 003 039 836 885 237 76;
  • 13) 0.000 003 039 836 885 237 76 × 2 = 0 + 0.000 006 079 673 770 475 52;
  • 14) 0.000 006 079 673 770 475 52 × 2 = 0 + 0.000 012 159 347 540 951 04;
  • 15) 0.000 012 159 347 540 951 04 × 2 = 0 + 0.000 024 318 695 081 902 08;
  • 16) 0.000 024 318 695 081 902 08 × 2 = 0 + 0.000 048 637 390 163 804 16;
  • 17) 0.000 048 637 390 163 804 16 × 2 = 0 + 0.000 097 274 780 327 608 32;
  • 18) 0.000 097 274 780 327 608 32 × 2 = 0 + 0.000 194 549 560 655 216 64;
  • 19) 0.000 194 549 560 655 216 64 × 2 = 0 + 0.000 389 099 121 310 433 28;
  • 20) 0.000 389 099 121 310 433 28 × 2 = 0 + 0.000 778 198 242 620 866 56;
  • 21) 0.000 778 198 242 620 866 56 × 2 = 0 + 0.001 556 396 485 241 733 12;
  • 22) 0.001 556 396 485 241 733 12 × 2 = 0 + 0.003 112 792 970 483 466 24;
  • 23) 0.003 112 792 970 483 466 24 × 2 = 0 + 0.006 225 585 940 966 932 48;
  • 24) 0.006 225 585 940 966 932 48 × 2 = 0 + 0.012 451 171 881 933 864 96;
  • 25) 0.012 451 171 881 933 864 96 × 2 = 0 + 0.024 902 343 763 867 729 92;
  • 26) 0.024 902 343 763 867 729 92 × 2 = 0 + 0.049 804 687 527 735 459 84;
  • 27) 0.049 804 687 527 735 459 84 × 2 = 0 + 0.099 609 375 055 470 919 68;
  • 28) 0.099 609 375 055 470 919 68 × 2 = 0 + 0.199 218 750 110 941 839 36;
  • 29) 0.199 218 750 110 941 839 36 × 2 = 0 + 0.398 437 500 221 883 678 72;
  • 30) 0.398 437 500 221 883 678 72 × 2 = 0 + 0.796 875 000 443 767 357 44;
  • 31) 0.796 875 000 443 767 357 44 × 2 = 1 + 0.593 750 000 887 534 714 88;
  • 32) 0.593 750 000 887 534 714 88 × 2 = 1 + 0.187 500 001 775 069 429 76;
  • 33) 0.187 500 001 775 069 429 76 × 2 = 0 + 0.375 000 003 550 138 859 52;
  • 34) 0.375 000 003 550 138 859 52 × 2 = 0 + 0.750 000 007 100 277 719 04;
  • 35) 0.750 000 007 100 277 719 04 × 2 = 1 + 0.500 000 014 200 555 438 08;
  • 36) 0.500 000 014 200 555 438 08 × 2 = 1 + 0.000 000 028 401 110 876 16;
  • 37) 0.000 000 028 401 110 876 16 × 2 = 0 + 0.000 000 056 802 221 752 32;
  • 38) 0.000 000 056 802 221 752 32 × 2 = 0 + 0.000 000 113 604 443 504 64;
  • 39) 0.000 000 113 604 443 504 64 × 2 = 0 + 0.000 000 227 208 887 009 28;
  • 40) 0.000 000 227 208 887 009 28 × 2 = 0 + 0.000 000 454 417 774 018 56;
  • 41) 0.000 000 454 417 774 018 56 × 2 = 0 + 0.000 000 908 835 548 037 12;
  • 42) 0.000 000 908 835 548 037 12 × 2 = 0 + 0.000 001 817 671 096 074 24;
  • 43) 0.000 001 817 671 096 074 24 × 2 = 0 + 0.000 003 635 342 192 148 48;
  • 44) 0.000 003 635 342 192 148 48 × 2 = 0 + 0.000 007 270 684 384 296 96;
  • 45) 0.000 007 270 684 384 296 96 × 2 = 0 + 0.000 014 541 368 768 593 92;
  • 46) 0.000 014 541 368 768 593 92 × 2 = 0 + 0.000 029 082 737 537 187 84;
  • 47) 0.000 029 082 737 537 187 84 × 2 = 0 + 0.000 058 165 475 074 375 68;
  • 48) 0.000 058 165 475 074 375 68 × 2 = 0 + 0.000 116 330 950 148 751 36;
  • 49) 0.000 116 330 950 148 751 36 × 2 = 0 + 0.000 232 661 900 297 502 72;
  • 50) 0.000 232 661 900 297 502 72 × 2 = 0 + 0.000 465 323 800 595 005 44;
  • 51) 0.000 465 323 800 595 005 44 × 2 = 0 + 0.000 930 647 601 190 010 88;
  • 52) 0.000 930 647 601 190 010 88 × 2 = 0 + 0.001 861 295 202 380 021 76;
  • 53) 0.001 861 295 202 380 021 76 × 2 = 0 + 0.003 722 590 404 760 043 52;
  • 54) 0.003 722 590 404 760 043 52 × 2 = 0 + 0.007 445 180 809 520 087 04;

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 06(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 06(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 06(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 06 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 77 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