-0.000 000 000 742 146 71 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 71(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
-0.000 000 000 742 146 71(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:

|-0.000 000 000 742 146 71| = 0.000 000 000 742 146 71


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: 0.000 000 000 742 146 71.

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) 0.000 000 000 742 146 71 × 2 = 0 + 0.000 000 001 484 293 42;
  • 2) 0.000 000 001 484 293 42 × 2 = 0 + 0.000 000 002 968 586 84;
  • 3) 0.000 000 002 968 586 84 × 2 = 0 + 0.000 000 005 937 173 68;
  • 4) 0.000 000 005 937 173 68 × 2 = 0 + 0.000 000 011 874 347 36;
  • 5) 0.000 000 011 874 347 36 × 2 = 0 + 0.000 000 023 748 694 72;
  • 6) 0.000 000 023 748 694 72 × 2 = 0 + 0.000 000 047 497 389 44;
  • 7) 0.000 000 047 497 389 44 × 2 = 0 + 0.000 000 094 994 778 88;
  • 8) 0.000 000 094 994 778 88 × 2 = 0 + 0.000 000 189 989 557 76;
  • 9) 0.000 000 189 989 557 76 × 2 = 0 + 0.000 000 379 979 115 52;
  • 10) 0.000 000 379 979 115 52 × 2 = 0 + 0.000 000 759 958 231 04;
  • 11) 0.000 000 759 958 231 04 × 2 = 0 + 0.000 001 519 916 462 08;
  • 12) 0.000 001 519 916 462 08 × 2 = 0 + 0.000 003 039 832 924 16;
  • 13) 0.000 003 039 832 924 16 × 2 = 0 + 0.000 006 079 665 848 32;
  • 14) 0.000 006 079 665 848 32 × 2 = 0 + 0.000 012 159 331 696 64;
  • 15) 0.000 012 159 331 696 64 × 2 = 0 + 0.000 024 318 663 393 28;
  • 16) 0.000 024 318 663 393 28 × 2 = 0 + 0.000 048 637 326 786 56;
  • 17) 0.000 048 637 326 786 56 × 2 = 0 + 0.000 097 274 653 573 12;
  • 18) 0.000 097 274 653 573 12 × 2 = 0 + 0.000 194 549 307 146 24;
  • 19) 0.000 194 549 307 146 24 × 2 = 0 + 0.000 389 098 614 292 48;
  • 20) 0.000 389 098 614 292 48 × 2 = 0 + 0.000 778 197 228 584 96;
  • 21) 0.000 778 197 228 584 96 × 2 = 0 + 0.001 556 394 457 169 92;
  • 22) 0.001 556 394 457 169 92 × 2 = 0 + 0.003 112 788 914 339 84;
  • 23) 0.003 112 788 914 339 84 × 2 = 0 + 0.006 225 577 828 679 68;
  • 24) 0.006 225 577 828 679 68 × 2 = 0 + 0.012 451 155 657 359 36;
  • 25) 0.012 451 155 657 359 36 × 2 = 0 + 0.024 902 311 314 718 72;
  • 26) 0.024 902 311 314 718 72 × 2 = 0 + 0.049 804 622 629 437 44;
  • 27) 0.049 804 622 629 437 44 × 2 = 0 + 0.099 609 245 258 874 88;
  • 28) 0.099 609 245 258 874 88 × 2 = 0 + 0.199 218 490 517 749 76;
  • 29) 0.199 218 490 517 749 76 × 2 = 0 + 0.398 436 981 035 499 52;
  • 30) 0.398 436 981 035 499 52 × 2 = 0 + 0.796 873 962 070 999 04;
  • 31) 0.796 873 962 070 999 04 × 2 = 1 + 0.593 747 924 141 998 08;
  • 32) 0.593 747 924 141 998 08 × 2 = 1 + 0.187 495 848 283 996 16;
  • 33) 0.187 495 848 283 996 16 × 2 = 0 + 0.374 991 696 567 992 32;
  • 34) 0.374 991 696 567 992 32 × 2 = 0 + 0.749 983 393 135 984 64;
  • 35) 0.749 983 393 135 984 64 × 2 = 1 + 0.499 966 786 271 969 28;
  • 36) 0.499 966 786 271 969 28 × 2 = 0 + 0.999 933 572 543 938 56;
  • 37) 0.999 933 572 543 938 56 × 2 = 1 + 0.999 867 145 087 877 12;
  • 38) 0.999 867 145 087 877 12 × 2 = 1 + 0.999 734 290 175 754 24;
  • 39) 0.999 734 290 175 754 24 × 2 = 1 + 0.999 468 580 351 508 48;
  • 40) 0.999 468 580 351 508 48 × 2 = 1 + 0.998 937 160 703 016 96;
  • 41) 0.998 937 160 703 016 96 × 2 = 1 + 0.997 874 321 406 033 92;
  • 42) 0.997 874 321 406 033 92 × 2 = 1 + 0.995 748 642 812 067 84;
  • 43) 0.995 748 642 812 067 84 × 2 = 1 + 0.991 497 285 624 135 68;
  • 44) 0.991 497 285 624 135 68 × 2 = 1 + 0.982 994 571 248 271 36;
  • 45) 0.982 994 571 248 271 36 × 2 = 1 + 0.965 989 142 496 542 72;
  • 46) 0.965 989 142 496 542 72 × 2 = 1 + 0.931 978 284 993 085 44;
  • 47) 0.931 978 284 993 085 44 × 2 = 1 + 0.863 956 569 986 170 88;
  • 48) 0.863 956 569 986 170 88 × 2 = 1 + 0.727 913 139 972 341 76;
  • 49) 0.727 913 139 972 341 76 × 2 = 1 + 0.455 826 279 944 683 52;
  • 50) 0.455 826 279 944 683 52 × 2 = 0 + 0.911 652 559 889 367 04;
  • 51) 0.911 652 559 889 367 04 × 2 = 1 + 0.823 305 119 778 734 08;
  • 52) 0.823 305 119 778 734 08 × 2 = 1 + 0.646 610 239 557 468 16;
  • 53) 0.646 610 239 557 468 16 × 2 = 1 + 0.293 220 479 114 936 32;
  • 54) 0.293 220 479 114 936 32 × 2 = 0 + 0.586 440 958 229 872 64;

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:


0.000 000 000 742 146 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 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:


0.000 000 000 742 146 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) × 20 =

1.1001 0111 1111 1111 1101 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 1101 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 1110 1110 =

100 1011 1111 1111 1110 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 1110 1110


Decimal number -0.000 000 000 742 146 71 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 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