-0.000 000 000 742 147 676 285 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 285(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
-0.000 000 000 742 147 676 285(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:

|-0.000 000 000 742 147 676 285| = 0.000 000 000 742 147 676 285


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: 0.000 000 000 742 147 676 285.

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) 0.000 000 000 742 147 676 285 × 2 = 0 + 0.000 000 001 484 295 352 57;
  • 2) 0.000 000 001 484 295 352 57 × 2 = 0 + 0.000 000 002 968 590 705 14;
  • 3) 0.000 000 002 968 590 705 14 × 2 = 0 + 0.000 000 005 937 181 410 28;
  • 4) 0.000 000 005 937 181 410 28 × 2 = 0 + 0.000 000 011 874 362 820 56;
  • 5) 0.000 000 011 874 362 820 56 × 2 = 0 + 0.000 000 023 748 725 641 12;
  • 6) 0.000 000 023 748 725 641 12 × 2 = 0 + 0.000 000 047 497 451 282 24;
  • 7) 0.000 000 047 497 451 282 24 × 2 = 0 + 0.000 000 094 994 902 564 48;
  • 8) 0.000 000 094 994 902 564 48 × 2 = 0 + 0.000 000 189 989 805 128 96;
  • 9) 0.000 000 189 989 805 128 96 × 2 = 0 + 0.000 000 379 979 610 257 92;
  • 10) 0.000 000 379 979 610 257 92 × 2 = 0 + 0.000 000 759 959 220 515 84;
  • 11) 0.000 000 759 959 220 515 84 × 2 = 0 + 0.000 001 519 918 441 031 68;
  • 12) 0.000 001 519 918 441 031 68 × 2 = 0 + 0.000 003 039 836 882 063 36;
  • 13) 0.000 003 039 836 882 063 36 × 2 = 0 + 0.000 006 079 673 764 126 72;
  • 14) 0.000 006 079 673 764 126 72 × 2 = 0 + 0.000 012 159 347 528 253 44;
  • 15) 0.000 012 159 347 528 253 44 × 2 = 0 + 0.000 024 318 695 056 506 88;
  • 16) 0.000 024 318 695 056 506 88 × 2 = 0 + 0.000 048 637 390 113 013 76;
  • 17) 0.000 048 637 390 113 013 76 × 2 = 0 + 0.000 097 274 780 226 027 52;
  • 18) 0.000 097 274 780 226 027 52 × 2 = 0 + 0.000 194 549 560 452 055 04;
  • 19) 0.000 194 549 560 452 055 04 × 2 = 0 + 0.000 389 099 120 904 110 08;
  • 20) 0.000 389 099 120 904 110 08 × 2 = 0 + 0.000 778 198 241 808 220 16;
  • 21) 0.000 778 198 241 808 220 16 × 2 = 0 + 0.001 556 396 483 616 440 32;
  • 22) 0.001 556 396 483 616 440 32 × 2 = 0 + 0.003 112 792 967 232 880 64;
  • 23) 0.003 112 792 967 232 880 64 × 2 = 0 + 0.006 225 585 934 465 761 28;
  • 24) 0.006 225 585 934 465 761 28 × 2 = 0 + 0.012 451 171 868 931 522 56;
  • 25) 0.012 451 171 868 931 522 56 × 2 = 0 + 0.024 902 343 737 863 045 12;
  • 26) 0.024 902 343 737 863 045 12 × 2 = 0 + 0.049 804 687 475 726 090 24;
  • 27) 0.049 804 687 475 726 090 24 × 2 = 0 + 0.099 609 374 951 452 180 48;
  • 28) 0.099 609 374 951 452 180 48 × 2 = 0 + 0.199 218 749 902 904 360 96;
  • 29) 0.199 218 749 902 904 360 96 × 2 = 0 + 0.398 437 499 805 808 721 92;
  • 30) 0.398 437 499 805 808 721 92 × 2 = 0 + 0.796 874 999 611 617 443 84;
  • 31) 0.796 874 999 611 617 443 84 × 2 = 1 + 0.593 749 999 223 234 887 68;
  • 32) 0.593 749 999 223 234 887 68 × 2 = 1 + 0.187 499 998 446 469 775 36;
  • 33) 0.187 499 998 446 469 775 36 × 2 = 0 + 0.374 999 996 892 939 550 72;
  • 34) 0.374 999 996 892 939 550 72 × 2 = 0 + 0.749 999 993 785 879 101 44;
  • 35) 0.749 999 993 785 879 101 44 × 2 = 1 + 0.499 999 987 571 758 202 88;
  • 36) 0.499 999 987 571 758 202 88 × 2 = 0 + 0.999 999 975 143 516 405 76;
  • 37) 0.999 999 975 143 516 405 76 × 2 = 1 + 0.999 999 950 287 032 811 52;
  • 38) 0.999 999 950 287 032 811 52 × 2 = 1 + 0.999 999 900 574 065 623 04;
  • 39) 0.999 999 900 574 065 623 04 × 2 = 1 + 0.999 999 801 148 131 246 08;
  • 40) 0.999 999 801 148 131 246 08 × 2 = 1 + 0.999 999 602 296 262 492 16;
  • 41) 0.999 999 602 296 262 492 16 × 2 = 1 + 0.999 999 204 592 524 984 32;
  • 42) 0.999 999 204 592 524 984 32 × 2 = 1 + 0.999 998 409 185 049 968 64;
  • 43) 0.999 998 409 185 049 968 64 × 2 = 1 + 0.999 996 818 370 099 937 28;
  • 44) 0.999 996 818 370 099 937 28 × 2 = 1 + 0.999 993 636 740 199 874 56;
  • 45) 0.999 993 636 740 199 874 56 × 2 = 1 + 0.999 987 273 480 399 749 12;
  • 46) 0.999 987 273 480 399 749 12 × 2 = 1 + 0.999 974 546 960 799 498 24;
  • 47) 0.999 974 546 960 799 498 24 × 2 = 1 + 0.999 949 093 921 598 996 48;
  • 48) 0.999 949 093 921 598 996 48 × 2 = 1 + 0.999 898 187 843 197 992 96;
  • 49) 0.999 898 187 843 197 992 96 × 2 = 1 + 0.999 796 375 686 395 985 92;
  • 50) 0.999 796 375 686 395 985 92 × 2 = 1 + 0.999 592 751 372 791 971 84;
  • 51) 0.999 592 751 372 791 971 84 × 2 = 1 + 0.999 185 502 745 583 943 68;
  • 52) 0.999 185 502 745 583 943 68 × 2 = 1 + 0.998 371 005 491 167 887 36;
  • 53) 0.998 371 005 491 167 887 36 × 2 = 1 + 0.996 742 010 982 335 774 72;
  • 54) 0.996 742 010 982 335 774 72 × 2 = 1 + 0.993 484 021 964 671 549 44;

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:


0.000 000 000 742 147 676 285(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 285(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:


0.000 000 000 742 147 676 285(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 -0.000 000 000 742 147 676 285 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