-0.000 000 000 742 147 673 63 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 63(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
-0.000 000 000 742 147 673 63(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:

|-0.000 000 000 742 147 673 63| = 0.000 000 000 742 147 673 63


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: 0.000 000 000 742 147 673 63.

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) 0.000 000 000 742 147 673 63 × 2 = 0 + 0.000 000 001 484 295 347 26;
  • 2) 0.000 000 001 484 295 347 26 × 2 = 0 + 0.000 000 002 968 590 694 52;
  • 3) 0.000 000 002 968 590 694 52 × 2 = 0 + 0.000 000 005 937 181 389 04;
  • 4) 0.000 000 005 937 181 389 04 × 2 = 0 + 0.000 000 011 874 362 778 08;
  • 5) 0.000 000 011 874 362 778 08 × 2 = 0 + 0.000 000 023 748 725 556 16;
  • 6) 0.000 000 023 748 725 556 16 × 2 = 0 + 0.000 000 047 497 451 112 32;
  • 7) 0.000 000 047 497 451 112 32 × 2 = 0 + 0.000 000 094 994 902 224 64;
  • 8) 0.000 000 094 994 902 224 64 × 2 = 0 + 0.000 000 189 989 804 449 28;
  • 9) 0.000 000 189 989 804 449 28 × 2 = 0 + 0.000 000 379 979 608 898 56;
  • 10) 0.000 000 379 979 608 898 56 × 2 = 0 + 0.000 000 759 959 217 797 12;
  • 11) 0.000 000 759 959 217 797 12 × 2 = 0 + 0.000 001 519 918 435 594 24;
  • 12) 0.000 001 519 918 435 594 24 × 2 = 0 + 0.000 003 039 836 871 188 48;
  • 13) 0.000 003 039 836 871 188 48 × 2 = 0 + 0.000 006 079 673 742 376 96;
  • 14) 0.000 006 079 673 742 376 96 × 2 = 0 + 0.000 012 159 347 484 753 92;
  • 15) 0.000 012 159 347 484 753 92 × 2 = 0 + 0.000 024 318 694 969 507 84;
  • 16) 0.000 024 318 694 969 507 84 × 2 = 0 + 0.000 048 637 389 939 015 68;
  • 17) 0.000 048 637 389 939 015 68 × 2 = 0 + 0.000 097 274 779 878 031 36;
  • 18) 0.000 097 274 779 878 031 36 × 2 = 0 + 0.000 194 549 559 756 062 72;
  • 19) 0.000 194 549 559 756 062 72 × 2 = 0 + 0.000 389 099 119 512 125 44;
  • 20) 0.000 389 099 119 512 125 44 × 2 = 0 + 0.000 778 198 239 024 250 88;
  • 21) 0.000 778 198 239 024 250 88 × 2 = 0 + 0.001 556 396 478 048 501 76;
  • 22) 0.001 556 396 478 048 501 76 × 2 = 0 + 0.003 112 792 956 097 003 52;
  • 23) 0.003 112 792 956 097 003 52 × 2 = 0 + 0.006 225 585 912 194 007 04;
  • 24) 0.006 225 585 912 194 007 04 × 2 = 0 + 0.012 451 171 824 388 014 08;
  • 25) 0.012 451 171 824 388 014 08 × 2 = 0 + 0.024 902 343 648 776 028 16;
  • 26) 0.024 902 343 648 776 028 16 × 2 = 0 + 0.049 804 687 297 552 056 32;
  • 27) 0.049 804 687 297 552 056 32 × 2 = 0 + 0.099 609 374 595 104 112 64;
  • 28) 0.099 609 374 595 104 112 64 × 2 = 0 + 0.199 218 749 190 208 225 28;
  • 29) 0.199 218 749 190 208 225 28 × 2 = 0 + 0.398 437 498 380 416 450 56;
  • 30) 0.398 437 498 380 416 450 56 × 2 = 0 + 0.796 874 996 760 832 901 12;
  • 31) 0.796 874 996 760 832 901 12 × 2 = 1 + 0.593 749 993 521 665 802 24;
  • 32) 0.593 749 993 521 665 802 24 × 2 = 1 + 0.187 499 987 043 331 604 48;
  • 33) 0.187 499 987 043 331 604 48 × 2 = 0 + 0.374 999 974 086 663 208 96;
  • 34) 0.374 999 974 086 663 208 96 × 2 = 0 + 0.749 999 948 173 326 417 92;
  • 35) 0.749 999 948 173 326 417 92 × 2 = 1 + 0.499 999 896 346 652 835 84;
  • 36) 0.499 999 896 346 652 835 84 × 2 = 0 + 0.999 999 792 693 305 671 68;
  • 37) 0.999 999 792 693 305 671 68 × 2 = 1 + 0.999 999 585 386 611 343 36;
  • 38) 0.999 999 585 386 611 343 36 × 2 = 1 + 0.999 999 170 773 222 686 72;
  • 39) 0.999 999 170 773 222 686 72 × 2 = 1 + 0.999 998 341 546 445 373 44;
  • 40) 0.999 998 341 546 445 373 44 × 2 = 1 + 0.999 996 683 092 890 746 88;
  • 41) 0.999 996 683 092 890 746 88 × 2 = 1 + 0.999 993 366 185 781 493 76;
  • 42) 0.999 993 366 185 781 493 76 × 2 = 1 + 0.999 986 732 371 562 987 52;
  • 43) 0.999 986 732 371 562 987 52 × 2 = 1 + 0.999 973 464 743 125 975 04;
  • 44) 0.999 973 464 743 125 975 04 × 2 = 1 + 0.999 946 929 486 251 950 08;
  • 45) 0.999 946 929 486 251 950 08 × 2 = 1 + 0.999 893 858 972 503 900 16;
  • 46) 0.999 893 858 972 503 900 16 × 2 = 1 + 0.999 787 717 945 007 800 32;
  • 47) 0.999 787 717 945 007 800 32 × 2 = 1 + 0.999 575 435 890 015 600 64;
  • 48) 0.999 575 435 890 015 600 64 × 2 = 1 + 0.999 150 871 780 031 201 28;
  • 49) 0.999 150 871 780 031 201 28 × 2 = 1 + 0.998 301 743 560 062 402 56;
  • 50) 0.998 301 743 560 062 402 56 × 2 = 1 + 0.996 603 487 120 124 805 12;
  • 51) 0.996 603 487 120 124 805 12 × 2 = 1 + 0.993 206 974 240 249 610 24;
  • 52) 0.993 206 974 240 249 610 24 × 2 = 1 + 0.986 413 948 480 499 220 48;
  • 53) 0.986 413 948 480 499 220 48 × 2 = 1 + 0.972 827 896 960 998 440 96;
  • 54) 0.972 827 896 960 998 440 96 × 2 = 1 + 0.945 655 793 921 996 881 92;

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:


0.000 000 000 742 147 673 63(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 63(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:


0.000 000 000 742 147 673 63(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 -0.000 000 000 742 147 673 63 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