-0.000 000 000 742 147 676 33 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 33(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 33(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 33| = 0.000 000 000 742 147 676 33


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 33.

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 33 × 2 = 0 + 0.000 000 001 484 295 352 66;
  • 2) 0.000 000 001 484 295 352 66 × 2 = 0 + 0.000 000 002 968 590 705 32;
  • 3) 0.000 000 002 968 590 705 32 × 2 = 0 + 0.000 000 005 937 181 410 64;
  • 4) 0.000 000 005 937 181 410 64 × 2 = 0 + 0.000 000 011 874 362 821 28;
  • 5) 0.000 000 011 874 362 821 28 × 2 = 0 + 0.000 000 023 748 725 642 56;
  • 6) 0.000 000 023 748 725 642 56 × 2 = 0 + 0.000 000 047 497 451 285 12;
  • 7) 0.000 000 047 497 451 285 12 × 2 = 0 + 0.000 000 094 994 902 570 24;
  • 8) 0.000 000 094 994 902 570 24 × 2 = 0 + 0.000 000 189 989 805 140 48;
  • 9) 0.000 000 189 989 805 140 48 × 2 = 0 + 0.000 000 379 979 610 280 96;
  • 10) 0.000 000 379 979 610 280 96 × 2 = 0 + 0.000 000 759 959 220 561 92;
  • 11) 0.000 000 759 959 220 561 92 × 2 = 0 + 0.000 001 519 918 441 123 84;
  • 12) 0.000 001 519 918 441 123 84 × 2 = 0 + 0.000 003 039 836 882 247 68;
  • 13) 0.000 003 039 836 882 247 68 × 2 = 0 + 0.000 006 079 673 764 495 36;
  • 14) 0.000 006 079 673 764 495 36 × 2 = 0 + 0.000 012 159 347 528 990 72;
  • 15) 0.000 012 159 347 528 990 72 × 2 = 0 + 0.000 024 318 695 057 981 44;
  • 16) 0.000 024 318 695 057 981 44 × 2 = 0 + 0.000 048 637 390 115 962 88;
  • 17) 0.000 048 637 390 115 962 88 × 2 = 0 + 0.000 097 274 780 231 925 76;
  • 18) 0.000 097 274 780 231 925 76 × 2 = 0 + 0.000 194 549 560 463 851 52;
  • 19) 0.000 194 549 560 463 851 52 × 2 = 0 + 0.000 389 099 120 927 703 04;
  • 20) 0.000 389 099 120 927 703 04 × 2 = 0 + 0.000 778 198 241 855 406 08;
  • 21) 0.000 778 198 241 855 406 08 × 2 = 0 + 0.001 556 396 483 710 812 16;
  • 22) 0.001 556 396 483 710 812 16 × 2 = 0 + 0.003 112 792 967 421 624 32;
  • 23) 0.003 112 792 967 421 624 32 × 2 = 0 + 0.006 225 585 934 843 248 64;
  • 24) 0.006 225 585 934 843 248 64 × 2 = 0 + 0.012 451 171 869 686 497 28;
  • 25) 0.012 451 171 869 686 497 28 × 2 = 0 + 0.024 902 343 739 372 994 56;
  • 26) 0.024 902 343 739 372 994 56 × 2 = 0 + 0.049 804 687 478 745 989 12;
  • 27) 0.049 804 687 478 745 989 12 × 2 = 0 + 0.099 609 374 957 491 978 24;
  • 28) 0.099 609 374 957 491 978 24 × 2 = 0 + 0.199 218 749 914 983 956 48;
  • 29) 0.199 218 749 914 983 956 48 × 2 = 0 + 0.398 437 499 829 967 912 96;
  • 30) 0.398 437 499 829 967 912 96 × 2 = 0 + 0.796 874 999 659 935 825 92;
  • 31) 0.796 874 999 659 935 825 92 × 2 = 1 + 0.593 749 999 319 871 651 84;
  • 32) 0.593 749 999 319 871 651 84 × 2 = 1 + 0.187 499 998 639 743 303 68;
  • 33) 0.187 499 998 639 743 303 68 × 2 = 0 + 0.374 999 997 279 486 607 36;
  • 34) 0.374 999 997 279 486 607 36 × 2 = 0 + 0.749 999 994 558 973 214 72;
  • 35) 0.749 999 994 558 973 214 72 × 2 = 1 + 0.499 999 989 117 946 429 44;
  • 36) 0.499 999 989 117 946 429 44 × 2 = 0 + 0.999 999 978 235 892 858 88;
  • 37) 0.999 999 978 235 892 858 88 × 2 = 1 + 0.999 999 956 471 785 717 76;
  • 38) 0.999 999 956 471 785 717 76 × 2 = 1 + 0.999 999 912 943 571 435 52;
  • 39) 0.999 999 912 943 571 435 52 × 2 = 1 + 0.999 999 825 887 142 871 04;
  • 40) 0.999 999 825 887 142 871 04 × 2 = 1 + 0.999 999 651 774 285 742 08;
  • 41) 0.999 999 651 774 285 742 08 × 2 = 1 + 0.999 999 303 548 571 484 16;
  • 42) 0.999 999 303 548 571 484 16 × 2 = 1 + 0.999 998 607 097 142 968 32;
  • 43) 0.999 998 607 097 142 968 32 × 2 = 1 + 0.999 997 214 194 285 936 64;
  • 44) 0.999 997 214 194 285 936 64 × 2 = 1 + 0.999 994 428 388 571 873 28;
  • 45) 0.999 994 428 388 571 873 28 × 2 = 1 + 0.999 988 856 777 143 746 56;
  • 46) 0.999 988 856 777 143 746 56 × 2 = 1 + 0.999 977 713 554 287 493 12;
  • 47) 0.999 977 713 554 287 493 12 × 2 = 1 + 0.999 955 427 108 574 986 24;
  • 48) 0.999 955 427 108 574 986 24 × 2 = 1 + 0.999 910 854 217 149 972 48;
  • 49) 0.999 910 854 217 149 972 48 × 2 = 1 + 0.999 821 708 434 299 944 96;
  • 50) 0.999 821 708 434 299 944 96 × 2 = 1 + 0.999 643 416 868 599 889 92;
  • 51) 0.999 643 416 868 599 889 92 × 2 = 1 + 0.999 286 833 737 199 779 84;
  • 52) 0.999 286 833 737 199 779 84 × 2 = 1 + 0.998 573 667 474 399 559 68;
  • 53) 0.998 573 667 474 399 559 68 × 2 = 1 + 0.997 147 334 948 799 119 36;
  • 54) 0.997 147 334 948 799 119 36 × 2 = 1 + 0.994 294 669 897 598 238 72;

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 33(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 33(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 33(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 33 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 675 64 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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