-0.000 000 000 742 147 676 475 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 475(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
-0.000 000 000 742 147 676 475(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:

|-0.000 000 000 742 147 676 475| = 0.000 000 000 742 147 676 475


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: 0.000 000 000 742 147 676 475.

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) 0.000 000 000 742 147 676 475 × 2 = 0 + 0.000 000 001 484 295 352 95;
  • 2) 0.000 000 001 484 295 352 95 × 2 = 0 + 0.000 000 002 968 590 705 9;
  • 3) 0.000 000 002 968 590 705 9 × 2 = 0 + 0.000 000 005 937 181 411 8;
  • 4) 0.000 000 005 937 181 411 8 × 2 = 0 + 0.000 000 011 874 362 823 6;
  • 5) 0.000 000 011 874 362 823 6 × 2 = 0 + 0.000 000 023 748 725 647 2;
  • 6) 0.000 000 023 748 725 647 2 × 2 = 0 + 0.000 000 047 497 451 294 4;
  • 7) 0.000 000 047 497 451 294 4 × 2 = 0 + 0.000 000 094 994 902 588 8;
  • 8) 0.000 000 094 994 902 588 8 × 2 = 0 + 0.000 000 189 989 805 177 6;
  • 9) 0.000 000 189 989 805 177 6 × 2 = 0 + 0.000 000 379 979 610 355 2;
  • 10) 0.000 000 379 979 610 355 2 × 2 = 0 + 0.000 000 759 959 220 710 4;
  • 11) 0.000 000 759 959 220 710 4 × 2 = 0 + 0.000 001 519 918 441 420 8;
  • 12) 0.000 001 519 918 441 420 8 × 2 = 0 + 0.000 003 039 836 882 841 6;
  • 13) 0.000 003 039 836 882 841 6 × 2 = 0 + 0.000 006 079 673 765 683 2;
  • 14) 0.000 006 079 673 765 683 2 × 2 = 0 + 0.000 012 159 347 531 366 4;
  • 15) 0.000 012 159 347 531 366 4 × 2 = 0 + 0.000 024 318 695 062 732 8;
  • 16) 0.000 024 318 695 062 732 8 × 2 = 0 + 0.000 048 637 390 125 465 6;
  • 17) 0.000 048 637 390 125 465 6 × 2 = 0 + 0.000 097 274 780 250 931 2;
  • 18) 0.000 097 274 780 250 931 2 × 2 = 0 + 0.000 194 549 560 501 862 4;
  • 19) 0.000 194 549 560 501 862 4 × 2 = 0 + 0.000 389 099 121 003 724 8;
  • 20) 0.000 389 099 121 003 724 8 × 2 = 0 + 0.000 778 198 242 007 449 6;
  • 21) 0.000 778 198 242 007 449 6 × 2 = 0 + 0.001 556 396 484 014 899 2;
  • 22) 0.001 556 396 484 014 899 2 × 2 = 0 + 0.003 112 792 968 029 798 4;
  • 23) 0.003 112 792 968 029 798 4 × 2 = 0 + 0.006 225 585 936 059 596 8;
  • 24) 0.006 225 585 936 059 596 8 × 2 = 0 + 0.012 451 171 872 119 193 6;
  • 25) 0.012 451 171 872 119 193 6 × 2 = 0 + 0.024 902 343 744 238 387 2;
  • 26) 0.024 902 343 744 238 387 2 × 2 = 0 + 0.049 804 687 488 476 774 4;
  • 27) 0.049 804 687 488 476 774 4 × 2 = 0 + 0.099 609 374 976 953 548 8;
  • 28) 0.099 609 374 976 953 548 8 × 2 = 0 + 0.199 218 749 953 907 097 6;
  • 29) 0.199 218 749 953 907 097 6 × 2 = 0 + 0.398 437 499 907 814 195 2;
  • 30) 0.398 437 499 907 814 195 2 × 2 = 0 + 0.796 874 999 815 628 390 4;
  • 31) 0.796 874 999 815 628 390 4 × 2 = 1 + 0.593 749 999 631 256 780 8;
  • 32) 0.593 749 999 631 256 780 8 × 2 = 1 + 0.187 499 999 262 513 561 6;
  • 33) 0.187 499 999 262 513 561 6 × 2 = 0 + 0.374 999 998 525 027 123 2;
  • 34) 0.374 999 998 525 027 123 2 × 2 = 0 + 0.749 999 997 050 054 246 4;
  • 35) 0.749 999 997 050 054 246 4 × 2 = 1 + 0.499 999 994 100 108 492 8;
  • 36) 0.499 999 994 100 108 492 8 × 2 = 0 + 0.999 999 988 200 216 985 6;
  • 37) 0.999 999 988 200 216 985 6 × 2 = 1 + 0.999 999 976 400 433 971 2;
  • 38) 0.999 999 976 400 433 971 2 × 2 = 1 + 0.999 999 952 800 867 942 4;
  • 39) 0.999 999 952 800 867 942 4 × 2 = 1 + 0.999 999 905 601 735 884 8;
  • 40) 0.999 999 905 601 735 884 8 × 2 = 1 + 0.999 999 811 203 471 769 6;
  • 41) 0.999 999 811 203 471 769 6 × 2 = 1 + 0.999 999 622 406 943 539 2;
  • 42) 0.999 999 622 406 943 539 2 × 2 = 1 + 0.999 999 244 813 887 078 4;
  • 43) 0.999 999 244 813 887 078 4 × 2 = 1 + 0.999 998 489 627 774 156 8;
  • 44) 0.999 998 489 627 774 156 8 × 2 = 1 + 0.999 996 979 255 548 313 6;
  • 45) 0.999 996 979 255 548 313 6 × 2 = 1 + 0.999 993 958 511 096 627 2;
  • 46) 0.999 993 958 511 096 627 2 × 2 = 1 + 0.999 987 917 022 193 254 4;
  • 47) 0.999 987 917 022 193 254 4 × 2 = 1 + 0.999 975 834 044 386 508 8;
  • 48) 0.999 975 834 044 386 508 8 × 2 = 1 + 0.999 951 668 088 773 017 6;
  • 49) 0.999 951 668 088 773 017 6 × 2 = 1 + 0.999 903 336 177 546 035 2;
  • 50) 0.999 903 336 177 546 035 2 × 2 = 1 + 0.999 806 672 355 092 070 4;
  • 51) 0.999 806 672 355 092 070 4 × 2 = 1 + 0.999 613 344 710 184 140 8;
  • 52) 0.999 613 344 710 184 140 8 × 2 = 1 + 0.999 226 689 420 368 281 6;
  • 53) 0.999 226 689 420 368 281 6 × 2 = 1 + 0.998 453 378 840 736 563 2;
  • 54) 0.998 453 378 840 736 563 2 × 2 = 1 + 0.996 906 757 681 473 126 4;

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:


0.000 000 000 742 147 676 475(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 475(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:


0.000 000 000 742 147 676 475(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 -0.000 000 000 742 147 676 475 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