-0.000 000 000 742 147 675 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 03(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
-0.000 000 000 742 147 675 03(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:

|-0.000 000 000 742 147 675 03| = 0.000 000 000 742 147 675 03


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: 0.000 000 000 742 147 675 03.

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) 0.000 000 000 742 147 675 03 × 2 = 0 + 0.000 000 001 484 295 350 06;
  • 2) 0.000 000 001 484 295 350 06 × 2 = 0 + 0.000 000 002 968 590 700 12;
  • 3) 0.000 000 002 968 590 700 12 × 2 = 0 + 0.000 000 005 937 181 400 24;
  • 4) 0.000 000 005 937 181 400 24 × 2 = 0 + 0.000 000 011 874 362 800 48;
  • 5) 0.000 000 011 874 362 800 48 × 2 = 0 + 0.000 000 023 748 725 600 96;
  • 6) 0.000 000 023 748 725 600 96 × 2 = 0 + 0.000 000 047 497 451 201 92;
  • 7) 0.000 000 047 497 451 201 92 × 2 = 0 + 0.000 000 094 994 902 403 84;
  • 8) 0.000 000 094 994 902 403 84 × 2 = 0 + 0.000 000 189 989 804 807 68;
  • 9) 0.000 000 189 989 804 807 68 × 2 = 0 + 0.000 000 379 979 609 615 36;
  • 10) 0.000 000 379 979 609 615 36 × 2 = 0 + 0.000 000 759 959 219 230 72;
  • 11) 0.000 000 759 959 219 230 72 × 2 = 0 + 0.000 001 519 918 438 461 44;
  • 12) 0.000 001 519 918 438 461 44 × 2 = 0 + 0.000 003 039 836 876 922 88;
  • 13) 0.000 003 039 836 876 922 88 × 2 = 0 + 0.000 006 079 673 753 845 76;
  • 14) 0.000 006 079 673 753 845 76 × 2 = 0 + 0.000 012 159 347 507 691 52;
  • 15) 0.000 012 159 347 507 691 52 × 2 = 0 + 0.000 024 318 695 015 383 04;
  • 16) 0.000 024 318 695 015 383 04 × 2 = 0 + 0.000 048 637 390 030 766 08;
  • 17) 0.000 048 637 390 030 766 08 × 2 = 0 + 0.000 097 274 780 061 532 16;
  • 18) 0.000 097 274 780 061 532 16 × 2 = 0 + 0.000 194 549 560 123 064 32;
  • 19) 0.000 194 549 560 123 064 32 × 2 = 0 + 0.000 389 099 120 246 128 64;
  • 20) 0.000 389 099 120 246 128 64 × 2 = 0 + 0.000 778 198 240 492 257 28;
  • 21) 0.000 778 198 240 492 257 28 × 2 = 0 + 0.001 556 396 480 984 514 56;
  • 22) 0.001 556 396 480 984 514 56 × 2 = 0 + 0.003 112 792 961 969 029 12;
  • 23) 0.003 112 792 961 969 029 12 × 2 = 0 + 0.006 225 585 923 938 058 24;
  • 24) 0.006 225 585 923 938 058 24 × 2 = 0 + 0.012 451 171 847 876 116 48;
  • 25) 0.012 451 171 847 876 116 48 × 2 = 0 + 0.024 902 343 695 752 232 96;
  • 26) 0.024 902 343 695 752 232 96 × 2 = 0 + 0.049 804 687 391 504 465 92;
  • 27) 0.049 804 687 391 504 465 92 × 2 = 0 + 0.099 609 374 783 008 931 84;
  • 28) 0.099 609 374 783 008 931 84 × 2 = 0 + 0.199 218 749 566 017 863 68;
  • 29) 0.199 218 749 566 017 863 68 × 2 = 0 + 0.398 437 499 132 035 727 36;
  • 30) 0.398 437 499 132 035 727 36 × 2 = 0 + 0.796 874 998 264 071 454 72;
  • 31) 0.796 874 998 264 071 454 72 × 2 = 1 + 0.593 749 996 528 142 909 44;
  • 32) 0.593 749 996 528 142 909 44 × 2 = 1 + 0.187 499 993 056 285 818 88;
  • 33) 0.187 499 993 056 285 818 88 × 2 = 0 + 0.374 999 986 112 571 637 76;
  • 34) 0.374 999 986 112 571 637 76 × 2 = 0 + 0.749 999 972 225 143 275 52;
  • 35) 0.749 999 972 225 143 275 52 × 2 = 1 + 0.499 999 944 450 286 551 04;
  • 36) 0.499 999 944 450 286 551 04 × 2 = 0 + 0.999 999 888 900 573 102 08;
  • 37) 0.999 999 888 900 573 102 08 × 2 = 1 + 0.999 999 777 801 146 204 16;
  • 38) 0.999 999 777 801 146 204 16 × 2 = 1 + 0.999 999 555 602 292 408 32;
  • 39) 0.999 999 555 602 292 408 32 × 2 = 1 + 0.999 999 111 204 584 816 64;
  • 40) 0.999 999 111 204 584 816 64 × 2 = 1 + 0.999 998 222 409 169 633 28;
  • 41) 0.999 998 222 409 169 633 28 × 2 = 1 + 0.999 996 444 818 339 266 56;
  • 42) 0.999 996 444 818 339 266 56 × 2 = 1 + 0.999 992 889 636 678 533 12;
  • 43) 0.999 992 889 636 678 533 12 × 2 = 1 + 0.999 985 779 273 357 066 24;
  • 44) 0.999 985 779 273 357 066 24 × 2 = 1 + 0.999 971 558 546 714 132 48;
  • 45) 0.999 971 558 546 714 132 48 × 2 = 1 + 0.999 943 117 093 428 264 96;
  • 46) 0.999 943 117 093 428 264 96 × 2 = 1 + 0.999 886 234 186 856 529 92;
  • 47) 0.999 886 234 186 856 529 92 × 2 = 1 + 0.999 772 468 373 713 059 84;
  • 48) 0.999 772 468 373 713 059 84 × 2 = 1 + 0.999 544 936 747 426 119 68;
  • 49) 0.999 544 936 747 426 119 68 × 2 = 1 + 0.999 089 873 494 852 239 36;
  • 50) 0.999 089 873 494 852 239 36 × 2 = 1 + 0.998 179 746 989 704 478 72;
  • 51) 0.998 179 746 989 704 478 72 × 2 = 1 + 0.996 359 493 979 408 957 44;
  • 52) 0.996 359 493 979 408 957 44 × 2 = 1 + 0.992 718 987 958 817 914 88;
  • 53) 0.992 718 987 958 817 914 88 × 2 = 1 + 0.985 437 975 917 635 829 76;
  • 54) 0.985 437 975 917 635 829 76 × 2 = 1 + 0.970 875 951 835 271 659 52;

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:


0.000 000 000 742 147 675 03(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 03(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:


0.000 000 000 742 147 675 03(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 -0.000 000 000 742 147 675 03 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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