-0.000 000 000 742 147 672 77 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 77(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 77(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 77| = 0.000 000 000 742 147 672 77


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 77.

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 77 × 2 = 0 + 0.000 000 001 484 295 345 54;
  • 2) 0.000 000 001 484 295 345 54 × 2 = 0 + 0.000 000 002 968 590 691 08;
  • 3) 0.000 000 002 968 590 691 08 × 2 = 0 + 0.000 000 005 937 181 382 16;
  • 4) 0.000 000 005 937 181 382 16 × 2 = 0 + 0.000 000 011 874 362 764 32;
  • 5) 0.000 000 011 874 362 764 32 × 2 = 0 + 0.000 000 023 748 725 528 64;
  • 6) 0.000 000 023 748 725 528 64 × 2 = 0 + 0.000 000 047 497 451 057 28;
  • 7) 0.000 000 047 497 451 057 28 × 2 = 0 + 0.000 000 094 994 902 114 56;
  • 8) 0.000 000 094 994 902 114 56 × 2 = 0 + 0.000 000 189 989 804 229 12;
  • 9) 0.000 000 189 989 804 229 12 × 2 = 0 + 0.000 000 379 979 608 458 24;
  • 10) 0.000 000 379 979 608 458 24 × 2 = 0 + 0.000 000 759 959 216 916 48;
  • 11) 0.000 000 759 959 216 916 48 × 2 = 0 + 0.000 001 519 918 433 832 96;
  • 12) 0.000 001 519 918 433 832 96 × 2 = 0 + 0.000 003 039 836 867 665 92;
  • 13) 0.000 003 039 836 867 665 92 × 2 = 0 + 0.000 006 079 673 735 331 84;
  • 14) 0.000 006 079 673 735 331 84 × 2 = 0 + 0.000 012 159 347 470 663 68;
  • 15) 0.000 012 159 347 470 663 68 × 2 = 0 + 0.000 024 318 694 941 327 36;
  • 16) 0.000 024 318 694 941 327 36 × 2 = 0 + 0.000 048 637 389 882 654 72;
  • 17) 0.000 048 637 389 882 654 72 × 2 = 0 + 0.000 097 274 779 765 309 44;
  • 18) 0.000 097 274 779 765 309 44 × 2 = 0 + 0.000 194 549 559 530 618 88;
  • 19) 0.000 194 549 559 530 618 88 × 2 = 0 + 0.000 389 099 119 061 237 76;
  • 20) 0.000 389 099 119 061 237 76 × 2 = 0 + 0.000 778 198 238 122 475 52;
  • 21) 0.000 778 198 238 122 475 52 × 2 = 0 + 0.001 556 396 476 244 951 04;
  • 22) 0.001 556 396 476 244 951 04 × 2 = 0 + 0.003 112 792 952 489 902 08;
  • 23) 0.003 112 792 952 489 902 08 × 2 = 0 + 0.006 225 585 904 979 804 16;
  • 24) 0.006 225 585 904 979 804 16 × 2 = 0 + 0.012 451 171 809 959 608 32;
  • 25) 0.012 451 171 809 959 608 32 × 2 = 0 + 0.024 902 343 619 919 216 64;
  • 26) 0.024 902 343 619 919 216 64 × 2 = 0 + 0.049 804 687 239 838 433 28;
  • 27) 0.049 804 687 239 838 433 28 × 2 = 0 + 0.099 609 374 479 676 866 56;
  • 28) 0.099 609 374 479 676 866 56 × 2 = 0 + 0.199 218 748 959 353 733 12;
  • 29) 0.199 218 748 959 353 733 12 × 2 = 0 + 0.398 437 497 918 707 466 24;
  • 30) 0.398 437 497 918 707 466 24 × 2 = 0 + 0.796 874 995 837 414 932 48;
  • 31) 0.796 874 995 837 414 932 48 × 2 = 1 + 0.593 749 991 674 829 864 96;
  • 32) 0.593 749 991 674 829 864 96 × 2 = 1 + 0.187 499 983 349 659 729 92;
  • 33) 0.187 499 983 349 659 729 92 × 2 = 0 + 0.374 999 966 699 319 459 84;
  • 34) 0.374 999 966 699 319 459 84 × 2 = 0 + 0.749 999 933 398 638 919 68;
  • 35) 0.749 999 933 398 638 919 68 × 2 = 1 + 0.499 999 866 797 277 839 36;
  • 36) 0.499 999 866 797 277 839 36 × 2 = 0 + 0.999 999 733 594 555 678 72;
  • 37) 0.999 999 733 594 555 678 72 × 2 = 1 + 0.999 999 467 189 111 357 44;
  • 38) 0.999 999 467 189 111 357 44 × 2 = 1 + 0.999 998 934 378 222 714 88;
  • 39) 0.999 998 934 378 222 714 88 × 2 = 1 + 0.999 997 868 756 445 429 76;
  • 40) 0.999 997 868 756 445 429 76 × 2 = 1 + 0.999 995 737 512 890 859 52;
  • 41) 0.999 995 737 512 890 859 52 × 2 = 1 + 0.999 991 475 025 781 719 04;
  • 42) 0.999 991 475 025 781 719 04 × 2 = 1 + 0.999 982 950 051 563 438 08;
  • 43) 0.999 982 950 051 563 438 08 × 2 = 1 + 0.999 965 900 103 126 876 16;
  • 44) 0.999 965 900 103 126 876 16 × 2 = 1 + 0.999 931 800 206 253 752 32;
  • 45) 0.999 931 800 206 253 752 32 × 2 = 1 + 0.999 863 600 412 507 504 64;
  • 46) 0.999 863 600 412 507 504 64 × 2 = 1 + 0.999 727 200 825 015 009 28;
  • 47) 0.999 727 200 825 015 009 28 × 2 = 1 + 0.999 454 401 650 030 018 56;
  • 48) 0.999 454 401 650 030 018 56 × 2 = 1 + 0.998 908 803 300 060 037 12;
  • 49) 0.998 908 803 300 060 037 12 × 2 = 1 + 0.997 817 606 600 120 074 24;
  • 50) 0.997 817 606 600 120 074 24 × 2 = 1 + 0.995 635 213 200 240 148 48;
  • 51) 0.995 635 213 200 240 148 48 × 2 = 1 + 0.991 270 426 400 480 296 96;
  • 52) 0.991 270 426 400 480 296 96 × 2 = 1 + 0.982 540 852 800 960 593 92;
  • 53) 0.982 540 852 800 960 593 92 × 2 = 1 + 0.965 081 705 601 921 187 84;
  • 54) 0.965 081 705 601 921 187 84 × 2 = 1 + 0.930 163 411 203 842 375 68;

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 77(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 77(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 77(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 77 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