-0.000 000 000 742 147 676 646 725 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 725 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
-0.000 000 000 742 147 676 646 725 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:

|-0.000 000 000 742 147 676 646 725 8| = 0.000 000 000 742 147 676 646 725 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: 0.000 000 000 742 147 676 646 725 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) 0.000 000 000 742 147 676 646 725 8 × 2 = 0 + 0.000 000 001 484 295 353 293 451 6;
  • 2) 0.000 000 001 484 295 353 293 451 6 × 2 = 0 + 0.000 000 002 968 590 706 586 903 2;
  • 3) 0.000 000 002 968 590 706 586 903 2 × 2 = 0 + 0.000 000 005 937 181 413 173 806 4;
  • 4) 0.000 000 005 937 181 413 173 806 4 × 2 = 0 + 0.000 000 011 874 362 826 347 612 8;
  • 5) 0.000 000 011 874 362 826 347 612 8 × 2 = 0 + 0.000 000 023 748 725 652 695 225 6;
  • 6) 0.000 000 023 748 725 652 695 225 6 × 2 = 0 + 0.000 000 047 497 451 305 390 451 2;
  • 7) 0.000 000 047 497 451 305 390 451 2 × 2 = 0 + 0.000 000 094 994 902 610 780 902 4;
  • 8) 0.000 000 094 994 902 610 780 902 4 × 2 = 0 + 0.000 000 189 989 805 221 561 804 8;
  • 9) 0.000 000 189 989 805 221 561 804 8 × 2 = 0 + 0.000 000 379 979 610 443 123 609 6;
  • 10) 0.000 000 379 979 610 443 123 609 6 × 2 = 0 + 0.000 000 759 959 220 886 247 219 2;
  • 11) 0.000 000 759 959 220 886 247 219 2 × 2 = 0 + 0.000 001 519 918 441 772 494 438 4;
  • 12) 0.000 001 519 918 441 772 494 438 4 × 2 = 0 + 0.000 003 039 836 883 544 988 876 8;
  • 13) 0.000 003 039 836 883 544 988 876 8 × 2 = 0 + 0.000 006 079 673 767 089 977 753 6;
  • 14) 0.000 006 079 673 767 089 977 753 6 × 2 = 0 + 0.000 012 159 347 534 179 955 507 2;
  • 15) 0.000 012 159 347 534 179 955 507 2 × 2 = 0 + 0.000 024 318 695 068 359 911 014 4;
  • 16) 0.000 024 318 695 068 359 911 014 4 × 2 = 0 + 0.000 048 637 390 136 719 822 028 8;
  • 17) 0.000 048 637 390 136 719 822 028 8 × 2 = 0 + 0.000 097 274 780 273 439 644 057 6;
  • 18) 0.000 097 274 780 273 439 644 057 6 × 2 = 0 + 0.000 194 549 560 546 879 288 115 2;
  • 19) 0.000 194 549 560 546 879 288 115 2 × 2 = 0 + 0.000 389 099 121 093 758 576 230 4;
  • 20) 0.000 389 099 121 093 758 576 230 4 × 2 = 0 + 0.000 778 198 242 187 517 152 460 8;
  • 21) 0.000 778 198 242 187 517 152 460 8 × 2 = 0 + 0.001 556 396 484 375 034 304 921 6;
  • 22) 0.001 556 396 484 375 034 304 921 6 × 2 = 0 + 0.003 112 792 968 750 068 609 843 2;
  • 23) 0.003 112 792 968 750 068 609 843 2 × 2 = 0 + 0.006 225 585 937 500 137 219 686 4;
  • 24) 0.006 225 585 937 500 137 219 686 4 × 2 = 0 + 0.012 451 171 875 000 274 439 372 8;
  • 25) 0.012 451 171 875 000 274 439 372 8 × 2 = 0 + 0.024 902 343 750 000 548 878 745 6;
  • 26) 0.024 902 343 750 000 548 878 745 6 × 2 = 0 + 0.049 804 687 500 001 097 757 491 2;
  • 27) 0.049 804 687 500 001 097 757 491 2 × 2 = 0 + 0.099 609 375 000 002 195 514 982 4;
  • 28) 0.099 609 375 000 002 195 514 982 4 × 2 = 0 + 0.199 218 750 000 004 391 029 964 8;
  • 29) 0.199 218 750 000 004 391 029 964 8 × 2 = 0 + 0.398 437 500 000 008 782 059 929 6;
  • 30) 0.398 437 500 000 008 782 059 929 6 × 2 = 0 + 0.796 875 000 000 017 564 119 859 2;
  • 31) 0.796 875 000 000 017 564 119 859 2 × 2 = 1 + 0.593 750 000 000 035 128 239 718 4;
  • 32) 0.593 750 000 000 035 128 239 718 4 × 2 = 1 + 0.187 500 000 000 070 256 479 436 8;
  • 33) 0.187 500 000 000 070 256 479 436 8 × 2 = 0 + 0.375 000 000 000 140 512 958 873 6;
  • 34) 0.375 000 000 000 140 512 958 873 6 × 2 = 0 + 0.750 000 000 000 281 025 917 747 2;
  • 35) 0.750 000 000 000 281 025 917 747 2 × 2 = 1 + 0.500 000 000 000 562 051 835 494 4;
  • 36) 0.500 000 000 000 562 051 835 494 4 × 2 = 1 + 0.000 000 000 001 124 103 670 988 8;
  • 37) 0.000 000 000 001 124 103 670 988 8 × 2 = 0 + 0.000 000 000 002 248 207 341 977 6;
  • 38) 0.000 000 000 002 248 207 341 977 6 × 2 = 0 + 0.000 000 000 004 496 414 683 955 2;
  • 39) 0.000 000 000 004 496 414 683 955 2 × 2 = 0 + 0.000 000 000 008 992 829 367 910 4;
  • 40) 0.000 000 000 008 992 829 367 910 4 × 2 = 0 + 0.000 000 000 017 985 658 735 820 8;
  • 41) 0.000 000 000 017 985 658 735 820 8 × 2 = 0 + 0.000 000 000 035 971 317 471 641 6;
  • 42) 0.000 000 000 035 971 317 471 641 6 × 2 = 0 + 0.000 000 000 071 942 634 943 283 2;
  • 43) 0.000 000 000 071 942 634 943 283 2 × 2 = 0 + 0.000 000 000 143 885 269 886 566 4;
  • 44) 0.000 000 000 143 885 269 886 566 4 × 2 = 0 + 0.000 000 000 287 770 539 773 132 8;
  • 45) 0.000 000 000 287 770 539 773 132 8 × 2 = 0 + 0.000 000 000 575 541 079 546 265 6;
  • 46) 0.000 000 000 575 541 079 546 265 6 × 2 = 0 + 0.000 000 001 151 082 159 092 531 2;
  • 47) 0.000 000 001 151 082 159 092 531 2 × 2 = 0 + 0.000 000 002 302 164 318 185 062 4;
  • 48) 0.000 000 002 302 164 318 185 062 4 × 2 = 0 + 0.000 000 004 604 328 636 370 124 8;
  • 49) 0.000 000 004 604 328 636 370 124 8 × 2 = 0 + 0.000 000 009 208 657 272 740 249 6;
  • 50) 0.000 000 009 208 657 272 740 249 6 × 2 = 0 + 0.000 000 018 417 314 545 480 499 2;
  • 51) 0.000 000 018 417 314 545 480 499 2 × 2 = 0 + 0.000 000 036 834 629 090 960 998 4;
  • 52) 0.000 000 036 834 629 090 960 998 4 × 2 = 0 + 0.000 000 073 669 258 181 921 996 8;
  • 53) 0.000 000 073 669 258 181 921 996 8 × 2 = 0 + 0.000 000 147 338 516 363 843 993 6;
  • 54) 0.000 000 147 338 516 363 843 993 6 × 2 = 0 + 0.000 000 294 677 032 727 687 987 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:


0.000 000 000 742 147 676 646 725 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 725 8(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:


0.000 000 000 742 147 676 646 725 8(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 -0.000 000 000 742 147 676 646 725 8 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