-0.000 000 000 742 147 673 05 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 05(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 05(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 05| = 0.000 000 000 742 147 673 05


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 05.

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 05 × 2 = 0 + 0.000 000 001 484 295 346 1;
  • 2) 0.000 000 001 484 295 346 1 × 2 = 0 + 0.000 000 002 968 590 692 2;
  • 3) 0.000 000 002 968 590 692 2 × 2 = 0 + 0.000 000 005 937 181 384 4;
  • 4) 0.000 000 005 937 181 384 4 × 2 = 0 + 0.000 000 011 874 362 768 8;
  • 5) 0.000 000 011 874 362 768 8 × 2 = 0 + 0.000 000 023 748 725 537 6;
  • 6) 0.000 000 023 748 725 537 6 × 2 = 0 + 0.000 000 047 497 451 075 2;
  • 7) 0.000 000 047 497 451 075 2 × 2 = 0 + 0.000 000 094 994 902 150 4;
  • 8) 0.000 000 094 994 902 150 4 × 2 = 0 + 0.000 000 189 989 804 300 8;
  • 9) 0.000 000 189 989 804 300 8 × 2 = 0 + 0.000 000 379 979 608 601 6;
  • 10) 0.000 000 379 979 608 601 6 × 2 = 0 + 0.000 000 759 959 217 203 2;
  • 11) 0.000 000 759 959 217 203 2 × 2 = 0 + 0.000 001 519 918 434 406 4;
  • 12) 0.000 001 519 918 434 406 4 × 2 = 0 + 0.000 003 039 836 868 812 8;
  • 13) 0.000 003 039 836 868 812 8 × 2 = 0 + 0.000 006 079 673 737 625 6;
  • 14) 0.000 006 079 673 737 625 6 × 2 = 0 + 0.000 012 159 347 475 251 2;
  • 15) 0.000 012 159 347 475 251 2 × 2 = 0 + 0.000 024 318 694 950 502 4;
  • 16) 0.000 024 318 694 950 502 4 × 2 = 0 + 0.000 048 637 389 901 004 8;
  • 17) 0.000 048 637 389 901 004 8 × 2 = 0 + 0.000 097 274 779 802 009 6;
  • 18) 0.000 097 274 779 802 009 6 × 2 = 0 + 0.000 194 549 559 604 019 2;
  • 19) 0.000 194 549 559 604 019 2 × 2 = 0 + 0.000 389 099 119 208 038 4;
  • 20) 0.000 389 099 119 208 038 4 × 2 = 0 + 0.000 778 198 238 416 076 8;
  • 21) 0.000 778 198 238 416 076 8 × 2 = 0 + 0.001 556 396 476 832 153 6;
  • 22) 0.001 556 396 476 832 153 6 × 2 = 0 + 0.003 112 792 953 664 307 2;
  • 23) 0.003 112 792 953 664 307 2 × 2 = 0 + 0.006 225 585 907 328 614 4;
  • 24) 0.006 225 585 907 328 614 4 × 2 = 0 + 0.012 451 171 814 657 228 8;
  • 25) 0.012 451 171 814 657 228 8 × 2 = 0 + 0.024 902 343 629 314 457 6;
  • 26) 0.024 902 343 629 314 457 6 × 2 = 0 + 0.049 804 687 258 628 915 2;
  • 27) 0.049 804 687 258 628 915 2 × 2 = 0 + 0.099 609 374 517 257 830 4;
  • 28) 0.099 609 374 517 257 830 4 × 2 = 0 + 0.199 218 749 034 515 660 8;
  • 29) 0.199 218 749 034 515 660 8 × 2 = 0 + 0.398 437 498 069 031 321 6;
  • 30) 0.398 437 498 069 031 321 6 × 2 = 0 + 0.796 874 996 138 062 643 2;
  • 31) 0.796 874 996 138 062 643 2 × 2 = 1 + 0.593 749 992 276 125 286 4;
  • 32) 0.593 749 992 276 125 286 4 × 2 = 1 + 0.187 499 984 552 250 572 8;
  • 33) 0.187 499 984 552 250 572 8 × 2 = 0 + 0.374 999 969 104 501 145 6;
  • 34) 0.374 999 969 104 501 145 6 × 2 = 0 + 0.749 999 938 209 002 291 2;
  • 35) 0.749 999 938 209 002 291 2 × 2 = 1 + 0.499 999 876 418 004 582 4;
  • 36) 0.499 999 876 418 004 582 4 × 2 = 0 + 0.999 999 752 836 009 164 8;
  • 37) 0.999 999 752 836 009 164 8 × 2 = 1 + 0.999 999 505 672 018 329 6;
  • 38) 0.999 999 505 672 018 329 6 × 2 = 1 + 0.999 999 011 344 036 659 2;
  • 39) 0.999 999 011 344 036 659 2 × 2 = 1 + 0.999 998 022 688 073 318 4;
  • 40) 0.999 998 022 688 073 318 4 × 2 = 1 + 0.999 996 045 376 146 636 8;
  • 41) 0.999 996 045 376 146 636 8 × 2 = 1 + 0.999 992 090 752 293 273 6;
  • 42) 0.999 992 090 752 293 273 6 × 2 = 1 + 0.999 984 181 504 586 547 2;
  • 43) 0.999 984 181 504 586 547 2 × 2 = 1 + 0.999 968 363 009 173 094 4;
  • 44) 0.999 968 363 009 173 094 4 × 2 = 1 + 0.999 936 726 018 346 188 8;
  • 45) 0.999 936 726 018 346 188 8 × 2 = 1 + 0.999 873 452 036 692 377 6;
  • 46) 0.999 873 452 036 692 377 6 × 2 = 1 + 0.999 746 904 073 384 755 2;
  • 47) 0.999 746 904 073 384 755 2 × 2 = 1 + 0.999 493 808 146 769 510 4;
  • 48) 0.999 493 808 146 769 510 4 × 2 = 1 + 0.998 987 616 293 539 020 8;
  • 49) 0.998 987 616 293 539 020 8 × 2 = 1 + 0.997 975 232 587 078 041 6;
  • 50) 0.997 975 232 587 078 041 6 × 2 = 1 + 0.995 950 465 174 156 083 2;
  • 51) 0.995 950 465 174 156 083 2 × 2 = 1 + 0.991 900 930 348 312 166 4;
  • 52) 0.991 900 930 348 312 166 4 × 2 = 1 + 0.983 801 860 696 624 332 8;
  • 53) 0.983 801 860 696 624 332 8 × 2 = 1 + 0.967 603 721 393 248 665 6;
  • 54) 0.967 603 721 393 248 665 6 × 2 = 1 + 0.935 207 442 786 497 331 2;

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 05(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 05(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 05(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 05 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