-0.000 000 000 742 147 676 245 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 245(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
-0.000 000 000 742 147 676 245(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:

|-0.000 000 000 742 147 676 245| = 0.000 000 000 742 147 676 245


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: 0.000 000 000 742 147 676 245.

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) 0.000 000 000 742 147 676 245 × 2 = 0 + 0.000 000 001 484 295 352 49;
  • 2) 0.000 000 001 484 295 352 49 × 2 = 0 + 0.000 000 002 968 590 704 98;
  • 3) 0.000 000 002 968 590 704 98 × 2 = 0 + 0.000 000 005 937 181 409 96;
  • 4) 0.000 000 005 937 181 409 96 × 2 = 0 + 0.000 000 011 874 362 819 92;
  • 5) 0.000 000 011 874 362 819 92 × 2 = 0 + 0.000 000 023 748 725 639 84;
  • 6) 0.000 000 023 748 725 639 84 × 2 = 0 + 0.000 000 047 497 451 279 68;
  • 7) 0.000 000 047 497 451 279 68 × 2 = 0 + 0.000 000 094 994 902 559 36;
  • 8) 0.000 000 094 994 902 559 36 × 2 = 0 + 0.000 000 189 989 805 118 72;
  • 9) 0.000 000 189 989 805 118 72 × 2 = 0 + 0.000 000 379 979 610 237 44;
  • 10) 0.000 000 379 979 610 237 44 × 2 = 0 + 0.000 000 759 959 220 474 88;
  • 11) 0.000 000 759 959 220 474 88 × 2 = 0 + 0.000 001 519 918 440 949 76;
  • 12) 0.000 001 519 918 440 949 76 × 2 = 0 + 0.000 003 039 836 881 899 52;
  • 13) 0.000 003 039 836 881 899 52 × 2 = 0 + 0.000 006 079 673 763 799 04;
  • 14) 0.000 006 079 673 763 799 04 × 2 = 0 + 0.000 012 159 347 527 598 08;
  • 15) 0.000 012 159 347 527 598 08 × 2 = 0 + 0.000 024 318 695 055 196 16;
  • 16) 0.000 024 318 695 055 196 16 × 2 = 0 + 0.000 048 637 390 110 392 32;
  • 17) 0.000 048 637 390 110 392 32 × 2 = 0 + 0.000 097 274 780 220 784 64;
  • 18) 0.000 097 274 780 220 784 64 × 2 = 0 + 0.000 194 549 560 441 569 28;
  • 19) 0.000 194 549 560 441 569 28 × 2 = 0 + 0.000 389 099 120 883 138 56;
  • 20) 0.000 389 099 120 883 138 56 × 2 = 0 + 0.000 778 198 241 766 277 12;
  • 21) 0.000 778 198 241 766 277 12 × 2 = 0 + 0.001 556 396 483 532 554 24;
  • 22) 0.001 556 396 483 532 554 24 × 2 = 0 + 0.003 112 792 967 065 108 48;
  • 23) 0.003 112 792 967 065 108 48 × 2 = 0 + 0.006 225 585 934 130 216 96;
  • 24) 0.006 225 585 934 130 216 96 × 2 = 0 + 0.012 451 171 868 260 433 92;
  • 25) 0.012 451 171 868 260 433 92 × 2 = 0 + 0.024 902 343 736 520 867 84;
  • 26) 0.024 902 343 736 520 867 84 × 2 = 0 + 0.049 804 687 473 041 735 68;
  • 27) 0.049 804 687 473 041 735 68 × 2 = 0 + 0.099 609 374 946 083 471 36;
  • 28) 0.099 609 374 946 083 471 36 × 2 = 0 + 0.199 218 749 892 166 942 72;
  • 29) 0.199 218 749 892 166 942 72 × 2 = 0 + 0.398 437 499 784 333 885 44;
  • 30) 0.398 437 499 784 333 885 44 × 2 = 0 + 0.796 874 999 568 667 770 88;
  • 31) 0.796 874 999 568 667 770 88 × 2 = 1 + 0.593 749 999 137 335 541 76;
  • 32) 0.593 749 999 137 335 541 76 × 2 = 1 + 0.187 499 998 274 671 083 52;
  • 33) 0.187 499 998 274 671 083 52 × 2 = 0 + 0.374 999 996 549 342 167 04;
  • 34) 0.374 999 996 549 342 167 04 × 2 = 0 + 0.749 999 993 098 684 334 08;
  • 35) 0.749 999 993 098 684 334 08 × 2 = 1 + 0.499 999 986 197 368 668 16;
  • 36) 0.499 999 986 197 368 668 16 × 2 = 0 + 0.999 999 972 394 737 336 32;
  • 37) 0.999 999 972 394 737 336 32 × 2 = 1 + 0.999 999 944 789 474 672 64;
  • 38) 0.999 999 944 789 474 672 64 × 2 = 1 + 0.999 999 889 578 949 345 28;
  • 39) 0.999 999 889 578 949 345 28 × 2 = 1 + 0.999 999 779 157 898 690 56;
  • 40) 0.999 999 779 157 898 690 56 × 2 = 1 + 0.999 999 558 315 797 381 12;
  • 41) 0.999 999 558 315 797 381 12 × 2 = 1 + 0.999 999 116 631 594 762 24;
  • 42) 0.999 999 116 631 594 762 24 × 2 = 1 + 0.999 998 233 263 189 524 48;
  • 43) 0.999 998 233 263 189 524 48 × 2 = 1 + 0.999 996 466 526 379 048 96;
  • 44) 0.999 996 466 526 379 048 96 × 2 = 1 + 0.999 992 933 052 758 097 92;
  • 45) 0.999 992 933 052 758 097 92 × 2 = 1 + 0.999 985 866 105 516 195 84;
  • 46) 0.999 985 866 105 516 195 84 × 2 = 1 + 0.999 971 732 211 032 391 68;
  • 47) 0.999 971 732 211 032 391 68 × 2 = 1 + 0.999 943 464 422 064 783 36;
  • 48) 0.999 943 464 422 064 783 36 × 2 = 1 + 0.999 886 928 844 129 566 72;
  • 49) 0.999 886 928 844 129 566 72 × 2 = 1 + 0.999 773 857 688 259 133 44;
  • 50) 0.999 773 857 688 259 133 44 × 2 = 1 + 0.999 547 715 376 518 266 88;
  • 51) 0.999 547 715 376 518 266 88 × 2 = 1 + 0.999 095 430 753 036 533 76;
  • 52) 0.999 095 430 753 036 533 76 × 2 = 1 + 0.998 190 861 506 073 067 52;
  • 53) 0.998 190 861 506 073 067 52 × 2 = 1 + 0.996 381 723 012 146 135 04;
  • 54) 0.996 381 723 012 146 135 04 × 2 = 1 + 0.992 763 446 024 292 270 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:


0.000 000 000 742 147 676 245(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 245(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:


0.000 000 000 742 147 676 245(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 -0.000 000 000 742 147 676 245 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