-0.000 000 000 742 147 676 37 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 37(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 37(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 37| = 0.000 000 000 742 147 676 37


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 37.

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 37 × 2 = 0 + 0.000 000 001 484 295 352 74;
  • 2) 0.000 000 001 484 295 352 74 × 2 = 0 + 0.000 000 002 968 590 705 48;
  • 3) 0.000 000 002 968 590 705 48 × 2 = 0 + 0.000 000 005 937 181 410 96;
  • 4) 0.000 000 005 937 181 410 96 × 2 = 0 + 0.000 000 011 874 362 821 92;
  • 5) 0.000 000 011 874 362 821 92 × 2 = 0 + 0.000 000 023 748 725 643 84;
  • 6) 0.000 000 023 748 725 643 84 × 2 = 0 + 0.000 000 047 497 451 287 68;
  • 7) 0.000 000 047 497 451 287 68 × 2 = 0 + 0.000 000 094 994 902 575 36;
  • 8) 0.000 000 094 994 902 575 36 × 2 = 0 + 0.000 000 189 989 805 150 72;
  • 9) 0.000 000 189 989 805 150 72 × 2 = 0 + 0.000 000 379 979 610 301 44;
  • 10) 0.000 000 379 979 610 301 44 × 2 = 0 + 0.000 000 759 959 220 602 88;
  • 11) 0.000 000 759 959 220 602 88 × 2 = 0 + 0.000 001 519 918 441 205 76;
  • 12) 0.000 001 519 918 441 205 76 × 2 = 0 + 0.000 003 039 836 882 411 52;
  • 13) 0.000 003 039 836 882 411 52 × 2 = 0 + 0.000 006 079 673 764 823 04;
  • 14) 0.000 006 079 673 764 823 04 × 2 = 0 + 0.000 012 159 347 529 646 08;
  • 15) 0.000 012 159 347 529 646 08 × 2 = 0 + 0.000 024 318 695 059 292 16;
  • 16) 0.000 024 318 695 059 292 16 × 2 = 0 + 0.000 048 637 390 118 584 32;
  • 17) 0.000 048 637 390 118 584 32 × 2 = 0 + 0.000 097 274 780 237 168 64;
  • 18) 0.000 097 274 780 237 168 64 × 2 = 0 + 0.000 194 549 560 474 337 28;
  • 19) 0.000 194 549 560 474 337 28 × 2 = 0 + 0.000 389 099 120 948 674 56;
  • 20) 0.000 389 099 120 948 674 56 × 2 = 0 + 0.000 778 198 241 897 349 12;
  • 21) 0.000 778 198 241 897 349 12 × 2 = 0 + 0.001 556 396 483 794 698 24;
  • 22) 0.001 556 396 483 794 698 24 × 2 = 0 + 0.003 112 792 967 589 396 48;
  • 23) 0.003 112 792 967 589 396 48 × 2 = 0 + 0.006 225 585 935 178 792 96;
  • 24) 0.006 225 585 935 178 792 96 × 2 = 0 + 0.012 451 171 870 357 585 92;
  • 25) 0.012 451 171 870 357 585 92 × 2 = 0 + 0.024 902 343 740 715 171 84;
  • 26) 0.024 902 343 740 715 171 84 × 2 = 0 + 0.049 804 687 481 430 343 68;
  • 27) 0.049 804 687 481 430 343 68 × 2 = 0 + 0.099 609 374 962 860 687 36;
  • 28) 0.099 609 374 962 860 687 36 × 2 = 0 + 0.199 218 749 925 721 374 72;
  • 29) 0.199 218 749 925 721 374 72 × 2 = 0 + 0.398 437 499 851 442 749 44;
  • 30) 0.398 437 499 851 442 749 44 × 2 = 0 + 0.796 874 999 702 885 498 88;
  • 31) 0.796 874 999 702 885 498 88 × 2 = 1 + 0.593 749 999 405 770 997 76;
  • 32) 0.593 749 999 405 770 997 76 × 2 = 1 + 0.187 499 998 811 541 995 52;
  • 33) 0.187 499 998 811 541 995 52 × 2 = 0 + 0.374 999 997 623 083 991 04;
  • 34) 0.374 999 997 623 083 991 04 × 2 = 0 + 0.749 999 995 246 167 982 08;
  • 35) 0.749 999 995 246 167 982 08 × 2 = 1 + 0.499 999 990 492 335 964 16;
  • 36) 0.499 999 990 492 335 964 16 × 2 = 0 + 0.999 999 980 984 671 928 32;
  • 37) 0.999 999 980 984 671 928 32 × 2 = 1 + 0.999 999 961 969 343 856 64;
  • 38) 0.999 999 961 969 343 856 64 × 2 = 1 + 0.999 999 923 938 687 713 28;
  • 39) 0.999 999 923 938 687 713 28 × 2 = 1 + 0.999 999 847 877 375 426 56;
  • 40) 0.999 999 847 877 375 426 56 × 2 = 1 + 0.999 999 695 754 750 853 12;
  • 41) 0.999 999 695 754 750 853 12 × 2 = 1 + 0.999 999 391 509 501 706 24;
  • 42) 0.999 999 391 509 501 706 24 × 2 = 1 + 0.999 998 783 019 003 412 48;
  • 43) 0.999 998 783 019 003 412 48 × 2 = 1 + 0.999 997 566 038 006 824 96;
  • 44) 0.999 997 566 038 006 824 96 × 2 = 1 + 0.999 995 132 076 013 649 92;
  • 45) 0.999 995 132 076 013 649 92 × 2 = 1 + 0.999 990 264 152 027 299 84;
  • 46) 0.999 990 264 152 027 299 84 × 2 = 1 + 0.999 980 528 304 054 599 68;
  • 47) 0.999 980 528 304 054 599 68 × 2 = 1 + 0.999 961 056 608 109 199 36;
  • 48) 0.999 961 056 608 109 199 36 × 2 = 1 + 0.999 922 113 216 218 398 72;
  • 49) 0.999 922 113 216 218 398 72 × 2 = 1 + 0.999 844 226 432 436 797 44;
  • 50) 0.999 844 226 432 436 797 44 × 2 = 1 + 0.999 688 452 864 873 594 88;
  • 51) 0.999 688 452 864 873 594 88 × 2 = 1 + 0.999 376 905 729 747 189 76;
  • 52) 0.999 376 905 729 747 189 76 × 2 = 1 + 0.998 753 811 459 494 379 52;
  • 53) 0.998 753 811 459 494 379 52 × 2 = 1 + 0.997 507 622 918 988 759 04;
  • 54) 0.997 507 622 918 988 759 04 × 2 = 1 + 0.995 015 245 837 977 518 08;

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 37(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 37(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 37(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 37 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 677 32 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