-0.000 000 000 742 147 672 84 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 84(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
-0.000 000 000 742 147 672 84(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:

|-0.000 000 000 742 147 672 84| = 0.000 000 000 742 147 672 84


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: 0.000 000 000 742 147 672 84.

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) 0.000 000 000 742 147 672 84 × 2 = 0 + 0.000 000 001 484 295 345 68;
  • 2) 0.000 000 001 484 295 345 68 × 2 = 0 + 0.000 000 002 968 590 691 36;
  • 3) 0.000 000 002 968 590 691 36 × 2 = 0 + 0.000 000 005 937 181 382 72;
  • 4) 0.000 000 005 937 181 382 72 × 2 = 0 + 0.000 000 011 874 362 765 44;
  • 5) 0.000 000 011 874 362 765 44 × 2 = 0 + 0.000 000 023 748 725 530 88;
  • 6) 0.000 000 023 748 725 530 88 × 2 = 0 + 0.000 000 047 497 451 061 76;
  • 7) 0.000 000 047 497 451 061 76 × 2 = 0 + 0.000 000 094 994 902 123 52;
  • 8) 0.000 000 094 994 902 123 52 × 2 = 0 + 0.000 000 189 989 804 247 04;
  • 9) 0.000 000 189 989 804 247 04 × 2 = 0 + 0.000 000 379 979 608 494 08;
  • 10) 0.000 000 379 979 608 494 08 × 2 = 0 + 0.000 000 759 959 216 988 16;
  • 11) 0.000 000 759 959 216 988 16 × 2 = 0 + 0.000 001 519 918 433 976 32;
  • 12) 0.000 001 519 918 433 976 32 × 2 = 0 + 0.000 003 039 836 867 952 64;
  • 13) 0.000 003 039 836 867 952 64 × 2 = 0 + 0.000 006 079 673 735 905 28;
  • 14) 0.000 006 079 673 735 905 28 × 2 = 0 + 0.000 012 159 347 471 810 56;
  • 15) 0.000 012 159 347 471 810 56 × 2 = 0 + 0.000 024 318 694 943 621 12;
  • 16) 0.000 024 318 694 943 621 12 × 2 = 0 + 0.000 048 637 389 887 242 24;
  • 17) 0.000 048 637 389 887 242 24 × 2 = 0 + 0.000 097 274 779 774 484 48;
  • 18) 0.000 097 274 779 774 484 48 × 2 = 0 + 0.000 194 549 559 548 968 96;
  • 19) 0.000 194 549 559 548 968 96 × 2 = 0 + 0.000 389 099 119 097 937 92;
  • 20) 0.000 389 099 119 097 937 92 × 2 = 0 + 0.000 778 198 238 195 875 84;
  • 21) 0.000 778 198 238 195 875 84 × 2 = 0 + 0.001 556 396 476 391 751 68;
  • 22) 0.001 556 396 476 391 751 68 × 2 = 0 + 0.003 112 792 952 783 503 36;
  • 23) 0.003 112 792 952 783 503 36 × 2 = 0 + 0.006 225 585 905 567 006 72;
  • 24) 0.006 225 585 905 567 006 72 × 2 = 0 + 0.012 451 171 811 134 013 44;
  • 25) 0.012 451 171 811 134 013 44 × 2 = 0 + 0.024 902 343 622 268 026 88;
  • 26) 0.024 902 343 622 268 026 88 × 2 = 0 + 0.049 804 687 244 536 053 76;
  • 27) 0.049 804 687 244 536 053 76 × 2 = 0 + 0.099 609 374 489 072 107 52;
  • 28) 0.099 609 374 489 072 107 52 × 2 = 0 + 0.199 218 748 978 144 215 04;
  • 29) 0.199 218 748 978 144 215 04 × 2 = 0 + 0.398 437 497 956 288 430 08;
  • 30) 0.398 437 497 956 288 430 08 × 2 = 0 + 0.796 874 995 912 576 860 16;
  • 31) 0.796 874 995 912 576 860 16 × 2 = 1 + 0.593 749 991 825 153 720 32;
  • 32) 0.593 749 991 825 153 720 32 × 2 = 1 + 0.187 499 983 650 307 440 64;
  • 33) 0.187 499 983 650 307 440 64 × 2 = 0 + 0.374 999 967 300 614 881 28;
  • 34) 0.374 999 967 300 614 881 28 × 2 = 0 + 0.749 999 934 601 229 762 56;
  • 35) 0.749 999 934 601 229 762 56 × 2 = 1 + 0.499 999 869 202 459 525 12;
  • 36) 0.499 999 869 202 459 525 12 × 2 = 0 + 0.999 999 738 404 919 050 24;
  • 37) 0.999 999 738 404 919 050 24 × 2 = 1 + 0.999 999 476 809 838 100 48;
  • 38) 0.999 999 476 809 838 100 48 × 2 = 1 + 0.999 998 953 619 676 200 96;
  • 39) 0.999 998 953 619 676 200 96 × 2 = 1 + 0.999 997 907 239 352 401 92;
  • 40) 0.999 997 907 239 352 401 92 × 2 = 1 + 0.999 995 814 478 704 803 84;
  • 41) 0.999 995 814 478 704 803 84 × 2 = 1 + 0.999 991 628 957 409 607 68;
  • 42) 0.999 991 628 957 409 607 68 × 2 = 1 + 0.999 983 257 914 819 215 36;
  • 43) 0.999 983 257 914 819 215 36 × 2 = 1 + 0.999 966 515 829 638 430 72;
  • 44) 0.999 966 515 829 638 430 72 × 2 = 1 + 0.999 933 031 659 276 861 44;
  • 45) 0.999 933 031 659 276 861 44 × 2 = 1 + 0.999 866 063 318 553 722 88;
  • 46) 0.999 866 063 318 553 722 88 × 2 = 1 + 0.999 732 126 637 107 445 76;
  • 47) 0.999 732 126 637 107 445 76 × 2 = 1 + 0.999 464 253 274 214 891 52;
  • 48) 0.999 464 253 274 214 891 52 × 2 = 1 + 0.998 928 506 548 429 783 04;
  • 49) 0.998 928 506 548 429 783 04 × 2 = 1 + 0.997 857 013 096 859 566 08;
  • 50) 0.997 857 013 096 859 566 08 × 2 = 1 + 0.995 714 026 193 719 132 16;
  • 51) 0.995 714 026 193 719 132 16 × 2 = 1 + 0.991 428 052 387 438 264 32;
  • 52) 0.991 428 052 387 438 264 32 × 2 = 1 + 0.982 856 104 774 876 528 64;
  • 53) 0.982 856 104 774 876 528 64 × 2 = 1 + 0.965 712 209 549 753 057 28;
  • 54) 0.965 712 209 549 753 057 28 × 2 = 1 + 0.931 424 419 099 506 114 56;

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:


0.000 000 000 742 147 672 84(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 672 84(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:


0.000 000 000 742 147 672 84(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 -0.000 000 000 742 147 672 84 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