-0.000 000 000 742 147 62 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 62(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 62(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 62| = 0.000 000 000 742 147 62


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 62.

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 62 × 2 = 0 + 0.000 000 001 484 295 24;
  • 2) 0.000 000 001 484 295 24 × 2 = 0 + 0.000 000 002 968 590 48;
  • 3) 0.000 000 002 968 590 48 × 2 = 0 + 0.000 000 005 937 180 96;
  • 4) 0.000 000 005 937 180 96 × 2 = 0 + 0.000 000 011 874 361 92;
  • 5) 0.000 000 011 874 361 92 × 2 = 0 + 0.000 000 023 748 723 84;
  • 6) 0.000 000 023 748 723 84 × 2 = 0 + 0.000 000 047 497 447 68;
  • 7) 0.000 000 047 497 447 68 × 2 = 0 + 0.000 000 094 994 895 36;
  • 8) 0.000 000 094 994 895 36 × 2 = 0 + 0.000 000 189 989 790 72;
  • 9) 0.000 000 189 989 790 72 × 2 = 0 + 0.000 000 379 979 581 44;
  • 10) 0.000 000 379 979 581 44 × 2 = 0 + 0.000 000 759 959 162 88;
  • 11) 0.000 000 759 959 162 88 × 2 = 0 + 0.000 001 519 918 325 76;
  • 12) 0.000 001 519 918 325 76 × 2 = 0 + 0.000 003 039 836 651 52;
  • 13) 0.000 003 039 836 651 52 × 2 = 0 + 0.000 006 079 673 303 04;
  • 14) 0.000 006 079 673 303 04 × 2 = 0 + 0.000 012 159 346 606 08;
  • 15) 0.000 012 159 346 606 08 × 2 = 0 + 0.000 024 318 693 212 16;
  • 16) 0.000 024 318 693 212 16 × 2 = 0 + 0.000 048 637 386 424 32;
  • 17) 0.000 048 637 386 424 32 × 2 = 0 + 0.000 097 274 772 848 64;
  • 18) 0.000 097 274 772 848 64 × 2 = 0 + 0.000 194 549 545 697 28;
  • 19) 0.000 194 549 545 697 28 × 2 = 0 + 0.000 389 099 091 394 56;
  • 20) 0.000 389 099 091 394 56 × 2 = 0 + 0.000 778 198 182 789 12;
  • 21) 0.000 778 198 182 789 12 × 2 = 0 + 0.001 556 396 365 578 24;
  • 22) 0.001 556 396 365 578 24 × 2 = 0 + 0.003 112 792 731 156 48;
  • 23) 0.003 112 792 731 156 48 × 2 = 0 + 0.006 225 585 462 312 96;
  • 24) 0.006 225 585 462 312 96 × 2 = 0 + 0.012 451 170 924 625 92;
  • 25) 0.012 451 170 924 625 92 × 2 = 0 + 0.024 902 341 849 251 84;
  • 26) 0.024 902 341 849 251 84 × 2 = 0 + 0.049 804 683 698 503 68;
  • 27) 0.049 804 683 698 503 68 × 2 = 0 + 0.099 609 367 397 007 36;
  • 28) 0.099 609 367 397 007 36 × 2 = 0 + 0.199 218 734 794 014 72;
  • 29) 0.199 218 734 794 014 72 × 2 = 0 + 0.398 437 469 588 029 44;
  • 30) 0.398 437 469 588 029 44 × 2 = 0 + 0.796 874 939 176 058 88;
  • 31) 0.796 874 939 176 058 88 × 2 = 1 + 0.593 749 878 352 117 76;
  • 32) 0.593 749 878 352 117 76 × 2 = 1 + 0.187 499 756 704 235 52;
  • 33) 0.187 499 756 704 235 52 × 2 = 0 + 0.374 999 513 408 471 04;
  • 34) 0.374 999 513 408 471 04 × 2 = 0 + 0.749 999 026 816 942 08;
  • 35) 0.749 999 026 816 942 08 × 2 = 1 + 0.499 998 053 633 884 16;
  • 36) 0.499 998 053 633 884 16 × 2 = 0 + 0.999 996 107 267 768 32;
  • 37) 0.999 996 107 267 768 32 × 2 = 1 + 0.999 992 214 535 536 64;
  • 38) 0.999 992 214 535 536 64 × 2 = 1 + 0.999 984 429 071 073 28;
  • 39) 0.999 984 429 071 073 28 × 2 = 1 + 0.999 968 858 142 146 56;
  • 40) 0.999 968 858 142 146 56 × 2 = 1 + 0.999 937 716 284 293 12;
  • 41) 0.999 937 716 284 293 12 × 2 = 1 + 0.999 875 432 568 586 24;
  • 42) 0.999 875 432 568 586 24 × 2 = 1 + 0.999 750 865 137 172 48;
  • 43) 0.999 750 865 137 172 48 × 2 = 1 + 0.999 501 730 274 344 96;
  • 44) 0.999 501 730 274 344 96 × 2 = 1 + 0.999 003 460 548 689 92;
  • 45) 0.999 003 460 548 689 92 × 2 = 1 + 0.998 006 921 097 379 84;
  • 46) 0.998 006 921 097 379 84 × 2 = 1 + 0.996 013 842 194 759 68;
  • 47) 0.996 013 842 194 759 68 × 2 = 1 + 0.992 027 684 389 519 36;
  • 48) 0.992 027 684 389 519 36 × 2 = 1 + 0.984 055 368 779 038 72;
  • 49) 0.984 055 368 779 038 72 × 2 = 1 + 0.968 110 737 558 077 44;
  • 50) 0.968 110 737 558 077 44 × 2 = 1 + 0.936 221 475 116 154 88;
  • 51) 0.936 221 475 116 154 88 × 2 = 1 + 0.872 442 950 232 309 76;
  • 52) 0.872 442 950 232 309 76 × 2 = 1 + 0.744 885 900 464 619 52;
  • 53) 0.744 885 900 464 619 52 × 2 = 1 + 0.489 771 800 929 239 04;
  • 54) 0.489 771 800 929 239 04 × 2 = 0 + 0.979 543 601 858 478 08;

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 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(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 62(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) × 20 =

1.1001 0111 1111 1111 1111 110(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 110


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 1110 =

100 1011 1111 1111 1111 1110


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 1110


Decimal number -0.000 000 000 742 147 62 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1110

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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