-0.000 000 000 742 147 676 35 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 35(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 35(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 35| = 0.000 000 000 742 147 676 35


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 35.

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 35 × 2 = 0 + 0.000 000 001 484 295 352 7;
  • 2) 0.000 000 001 484 295 352 7 × 2 = 0 + 0.000 000 002 968 590 705 4;
  • 3) 0.000 000 002 968 590 705 4 × 2 = 0 + 0.000 000 005 937 181 410 8;
  • 4) 0.000 000 005 937 181 410 8 × 2 = 0 + 0.000 000 011 874 362 821 6;
  • 5) 0.000 000 011 874 362 821 6 × 2 = 0 + 0.000 000 023 748 725 643 2;
  • 6) 0.000 000 023 748 725 643 2 × 2 = 0 + 0.000 000 047 497 451 286 4;
  • 7) 0.000 000 047 497 451 286 4 × 2 = 0 + 0.000 000 094 994 902 572 8;
  • 8) 0.000 000 094 994 902 572 8 × 2 = 0 + 0.000 000 189 989 805 145 6;
  • 9) 0.000 000 189 989 805 145 6 × 2 = 0 + 0.000 000 379 979 610 291 2;
  • 10) 0.000 000 379 979 610 291 2 × 2 = 0 + 0.000 000 759 959 220 582 4;
  • 11) 0.000 000 759 959 220 582 4 × 2 = 0 + 0.000 001 519 918 441 164 8;
  • 12) 0.000 001 519 918 441 164 8 × 2 = 0 + 0.000 003 039 836 882 329 6;
  • 13) 0.000 003 039 836 882 329 6 × 2 = 0 + 0.000 006 079 673 764 659 2;
  • 14) 0.000 006 079 673 764 659 2 × 2 = 0 + 0.000 012 159 347 529 318 4;
  • 15) 0.000 012 159 347 529 318 4 × 2 = 0 + 0.000 024 318 695 058 636 8;
  • 16) 0.000 024 318 695 058 636 8 × 2 = 0 + 0.000 048 637 390 117 273 6;
  • 17) 0.000 048 637 390 117 273 6 × 2 = 0 + 0.000 097 274 780 234 547 2;
  • 18) 0.000 097 274 780 234 547 2 × 2 = 0 + 0.000 194 549 560 469 094 4;
  • 19) 0.000 194 549 560 469 094 4 × 2 = 0 + 0.000 389 099 120 938 188 8;
  • 20) 0.000 389 099 120 938 188 8 × 2 = 0 + 0.000 778 198 241 876 377 6;
  • 21) 0.000 778 198 241 876 377 6 × 2 = 0 + 0.001 556 396 483 752 755 2;
  • 22) 0.001 556 396 483 752 755 2 × 2 = 0 + 0.003 112 792 967 505 510 4;
  • 23) 0.003 112 792 967 505 510 4 × 2 = 0 + 0.006 225 585 935 011 020 8;
  • 24) 0.006 225 585 935 011 020 8 × 2 = 0 + 0.012 451 171 870 022 041 6;
  • 25) 0.012 451 171 870 022 041 6 × 2 = 0 + 0.024 902 343 740 044 083 2;
  • 26) 0.024 902 343 740 044 083 2 × 2 = 0 + 0.049 804 687 480 088 166 4;
  • 27) 0.049 804 687 480 088 166 4 × 2 = 0 + 0.099 609 374 960 176 332 8;
  • 28) 0.099 609 374 960 176 332 8 × 2 = 0 + 0.199 218 749 920 352 665 6;
  • 29) 0.199 218 749 920 352 665 6 × 2 = 0 + 0.398 437 499 840 705 331 2;
  • 30) 0.398 437 499 840 705 331 2 × 2 = 0 + 0.796 874 999 681 410 662 4;
  • 31) 0.796 874 999 681 410 662 4 × 2 = 1 + 0.593 749 999 362 821 324 8;
  • 32) 0.593 749 999 362 821 324 8 × 2 = 1 + 0.187 499 998 725 642 649 6;
  • 33) 0.187 499 998 725 642 649 6 × 2 = 0 + 0.374 999 997 451 285 299 2;
  • 34) 0.374 999 997 451 285 299 2 × 2 = 0 + 0.749 999 994 902 570 598 4;
  • 35) 0.749 999 994 902 570 598 4 × 2 = 1 + 0.499 999 989 805 141 196 8;
  • 36) 0.499 999 989 805 141 196 8 × 2 = 0 + 0.999 999 979 610 282 393 6;
  • 37) 0.999 999 979 610 282 393 6 × 2 = 1 + 0.999 999 959 220 564 787 2;
  • 38) 0.999 999 959 220 564 787 2 × 2 = 1 + 0.999 999 918 441 129 574 4;
  • 39) 0.999 999 918 441 129 574 4 × 2 = 1 + 0.999 999 836 882 259 148 8;
  • 40) 0.999 999 836 882 259 148 8 × 2 = 1 + 0.999 999 673 764 518 297 6;
  • 41) 0.999 999 673 764 518 297 6 × 2 = 1 + 0.999 999 347 529 036 595 2;
  • 42) 0.999 999 347 529 036 595 2 × 2 = 1 + 0.999 998 695 058 073 190 4;
  • 43) 0.999 998 695 058 073 190 4 × 2 = 1 + 0.999 997 390 116 146 380 8;
  • 44) 0.999 997 390 116 146 380 8 × 2 = 1 + 0.999 994 780 232 292 761 6;
  • 45) 0.999 994 780 232 292 761 6 × 2 = 1 + 0.999 989 560 464 585 523 2;
  • 46) 0.999 989 560 464 585 523 2 × 2 = 1 + 0.999 979 120 929 171 046 4;
  • 47) 0.999 979 120 929 171 046 4 × 2 = 1 + 0.999 958 241 858 342 092 8;
  • 48) 0.999 958 241 858 342 092 8 × 2 = 1 + 0.999 916 483 716 684 185 6;
  • 49) 0.999 916 483 716 684 185 6 × 2 = 1 + 0.999 832 967 433 368 371 2;
  • 50) 0.999 832 967 433 368 371 2 × 2 = 1 + 0.999 665 934 866 736 742 4;
  • 51) 0.999 665 934 866 736 742 4 × 2 = 1 + 0.999 331 869 733 473 484 8;
  • 52) 0.999 331 869 733 473 484 8 × 2 = 1 + 0.998 663 739 466 946 969 6;
  • 53) 0.998 663 739 466 946 969 6 × 2 = 1 + 0.997 327 478 933 893 939 2;
  • 54) 0.997 327 478 933 893 939 2 × 2 = 1 + 0.994 654 957 867 787 878 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 35(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 35(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 35(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 35 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