-0.000 000 000 742 147 666 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 666 8(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
-0.000 000 000 742 147 666 8(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:

|-0.000 000 000 742 147 666 8| = 0.000 000 000 742 147 666 8


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: 0.000 000 000 742 147 666 8.

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) 0.000 000 000 742 147 666 8 × 2 = 0 + 0.000 000 001 484 295 333 6;
  • 2) 0.000 000 001 484 295 333 6 × 2 = 0 + 0.000 000 002 968 590 667 2;
  • 3) 0.000 000 002 968 590 667 2 × 2 = 0 + 0.000 000 005 937 181 334 4;
  • 4) 0.000 000 005 937 181 334 4 × 2 = 0 + 0.000 000 011 874 362 668 8;
  • 5) 0.000 000 011 874 362 668 8 × 2 = 0 + 0.000 000 023 748 725 337 6;
  • 6) 0.000 000 023 748 725 337 6 × 2 = 0 + 0.000 000 047 497 450 675 2;
  • 7) 0.000 000 047 497 450 675 2 × 2 = 0 + 0.000 000 094 994 901 350 4;
  • 8) 0.000 000 094 994 901 350 4 × 2 = 0 + 0.000 000 189 989 802 700 8;
  • 9) 0.000 000 189 989 802 700 8 × 2 = 0 + 0.000 000 379 979 605 401 6;
  • 10) 0.000 000 379 979 605 401 6 × 2 = 0 + 0.000 000 759 959 210 803 2;
  • 11) 0.000 000 759 959 210 803 2 × 2 = 0 + 0.000 001 519 918 421 606 4;
  • 12) 0.000 001 519 918 421 606 4 × 2 = 0 + 0.000 003 039 836 843 212 8;
  • 13) 0.000 003 039 836 843 212 8 × 2 = 0 + 0.000 006 079 673 686 425 6;
  • 14) 0.000 006 079 673 686 425 6 × 2 = 0 + 0.000 012 159 347 372 851 2;
  • 15) 0.000 012 159 347 372 851 2 × 2 = 0 + 0.000 024 318 694 745 702 4;
  • 16) 0.000 024 318 694 745 702 4 × 2 = 0 + 0.000 048 637 389 491 404 8;
  • 17) 0.000 048 637 389 491 404 8 × 2 = 0 + 0.000 097 274 778 982 809 6;
  • 18) 0.000 097 274 778 982 809 6 × 2 = 0 + 0.000 194 549 557 965 619 2;
  • 19) 0.000 194 549 557 965 619 2 × 2 = 0 + 0.000 389 099 115 931 238 4;
  • 20) 0.000 389 099 115 931 238 4 × 2 = 0 + 0.000 778 198 231 862 476 8;
  • 21) 0.000 778 198 231 862 476 8 × 2 = 0 + 0.001 556 396 463 724 953 6;
  • 22) 0.001 556 396 463 724 953 6 × 2 = 0 + 0.003 112 792 927 449 907 2;
  • 23) 0.003 112 792 927 449 907 2 × 2 = 0 + 0.006 225 585 854 899 814 4;
  • 24) 0.006 225 585 854 899 814 4 × 2 = 0 + 0.012 451 171 709 799 628 8;
  • 25) 0.012 451 171 709 799 628 8 × 2 = 0 + 0.024 902 343 419 599 257 6;
  • 26) 0.024 902 343 419 599 257 6 × 2 = 0 + 0.049 804 686 839 198 515 2;
  • 27) 0.049 804 686 839 198 515 2 × 2 = 0 + 0.099 609 373 678 397 030 4;
  • 28) 0.099 609 373 678 397 030 4 × 2 = 0 + 0.199 218 747 356 794 060 8;
  • 29) 0.199 218 747 356 794 060 8 × 2 = 0 + 0.398 437 494 713 588 121 6;
  • 30) 0.398 437 494 713 588 121 6 × 2 = 0 + 0.796 874 989 427 176 243 2;
  • 31) 0.796 874 989 427 176 243 2 × 2 = 1 + 0.593 749 978 854 352 486 4;
  • 32) 0.593 749 978 854 352 486 4 × 2 = 1 + 0.187 499 957 708 704 972 8;
  • 33) 0.187 499 957 708 704 972 8 × 2 = 0 + 0.374 999 915 417 409 945 6;
  • 34) 0.374 999 915 417 409 945 6 × 2 = 0 + 0.749 999 830 834 819 891 2;
  • 35) 0.749 999 830 834 819 891 2 × 2 = 1 + 0.499 999 661 669 639 782 4;
  • 36) 0.499 999 661 669 639 782 4 × 2 = 0 + 0.999 999 323 339 279 564 8;
  • 37) 0.999 999 323 339 279 564 8 × 2 = 1 + 0.999 998 646 678 559 129 6;
  • 38) 0.999 998 646 678 559 129 6 × 2 = 1 + 0.999 997 293 357 118 259 2;
  • 39) 0.999 997 293 357 118 259 2 × 2 = 1 + 0.999 994 586 714 236 518 4;
  • 40) 0.999 994 586 714 236 518 4 × 2 = 1 + 0.999 989 173 428 473 036 8;
  • 41) 0.999 989 173 428 473 036 8 × 2 = 1 + 0.999 978 346 856 946 073 6;
  • 42) 0.999 978 346 856 946 073 6 × 2 = 1 + 0.999 956 693 713 892 147 2;
  • 43) 0.999 956 693 713 892 147 2 × 2 = 1 + 0.999 913 387 427 784 294 4;
  • 44) 0.999 913 387 427 784 294 4 × 2 = 1 + 0.999 826 774 855 568 588 8;
  • 45) 0.999 826 774 855 568 588 8 × 2 = 1 + 0.999 653 549 711 137 177 6;
  • 46) 0.999 653 549 711 137 177 6 × 2 = 1 + 0.999 307 099 422 274 355 2;
  • 47) 0.999 307 099 422 274 355 2 × 2 = 1 + 0.998 614 198 844 548 710 4;
  • 48) 0.998 614 198 844 548 710 4 × 2 = 1 + 0.997 228 397 689 097 420 8;
  • 49) 0.997 228 397 689 097 420 8 × 2 = 1 + 0.994 456 795 378 194 841 6;
  • 50) 0.994 456 795 378 194 841 6 × 2 = 1 + 0.988 913 590 756 389 683 2;
  • 51) 0.988 913 590 756 389 683 2 × 2 = 1 + 0.977 827 181 512 779 366 4;
  • 52) 0.977 827 181 512 779 366 4 × 2 = 1 + 0.955 654 363 025 558 732 8;
  • 53) 0.955 654 363 025 558 732 8 × 2 = 1 + 0.911 308 726 051 117 465 6;
  • 54) 0.911 308 726 051 117 465 6 × 2 = 1 + 0.822 617 452 102 234 931 2;

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:


0.000 000 000 742 147 666 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 666 8(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:


0.000 000 000 742 147 666 8(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 -0.000 000 000 742 147 666 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 673 3 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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