-0.000 000 000 742 147 675 83 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 83(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
-0.000 000 000 742 147 675 83(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:

|-0.000 000 000 742 147 675 83| = 0.000 000 000 742 147 675 83


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: 0.000 000 000 742 147 675 83.

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) 0.000 000 000 742 147 675 83 × 2 = 0 + 0.000 000 001 484 295 351 66;
  • 2) 0.000 000 001 484 295 351 66 × 2 = 0 + 0.000 000 002 968 590 703 32;
  • 3) 0.000 000 002 968 590 703 32 × 2 = 0 + 0.000 000 005 937 181 406 64;
  • 4) 0.000 000 005 937 181 406 64 × 2 = 0 + 0.000 000 011 874 362 813 28;
  • 5) 0.000 000 011 874 362 813 28 × 2 = 0 + 0.000 000 023 748 725 626 56;
  • 6) 0.000 000 023 748 725 626 56 × 2 = 0 + 0.000 000 047 497 451 253 12;
  • 7) 0.000 000 047 497 451 253 12 × 2 = 0 + 0.000 000 094 994 902 506 24;
  • 8) 0.000 000 094 994 902 506 24 × 2 = 0 + 0.000 000 189 989 805 012 48;
  • 9) 0.000 000 189 989 805 012 48 × 2 = 0 + 0.000 000 379 979 610 024 96;
  • 10) 0.000 000 379 979 610 024 96 × 2 = 0 + 0.000 000 759 959 220 049 92;
  • 11) 0.000 000 759 959 220 049 92 × 2 = 0 + 0.000 001 519 918 440 099 84;
  • 12) 0.000 001 519 918 440 099 84 × 2 = 0 + 0.000 003 039 836 880 199 68;
  • 13) 0.000 003 039 836 880 199 68 × 2 = 0 + 0.000 006 079 673 760 399 36;
  • 14) 0.000 006 079 673 760 399 36 × 2 = 0 + 0.000 012 159 347 520 798 72;
  • 15) 0.000 012 159 347 520 798 72 × 2 = 0 + 0.000 024 318 695 041 597 44;
  • 16) 0.000 024 318 695 041 597 44 × 2 = 0 + 0.000 048 637 390 083 194 88;
  • 17) 0.000 048 637 390 083 194 88 × 2 = 0 + 0.000 097 274 780 166 389 76;
  • 18) 0.000 097 274 780 166 389 76 × 2 = 0 + 0.000 194 549 560 332 779 52;
  • 19) 0.000 194 549 560 332 779 52 × 2 = 0 + 0.000 389 099 120 665 559 04;
  • 20) 0.000 389 099 120 665 559 04 × 2 = 0 + 0.000 778 198 241 331 118 08;
  • 21) 0.000 778 198 241 331 118 08 × 2 = 0 + 0.001 556 396 482 662 236 16;
  • 22) 0.001 556 396 482 662 236 16 × 2 = 0 + 0.003 112 792 965 324 472 32;
  • 23) 0.003 112 792 965 324 472 32 × 2 = 0 + 0.006 225 585 930 648 944 64;
  • 24) 0.006 225 585 930 648 944 64 × 2 = 0 + 0.012 451 171 861 297 889 28;
  • 25) 0.012 451 171 861 297 889 28 × 2 = 0 + 0.024 902 343 722 595 778 56;
  • 26) 0.024 902 343 722 595 778 56 × 2 = 0 + 0.049 804 687 445 191 557 12;
  • 27) 0.049 804 687 445 191 557 12 × 2 = 0 + 0.099 609 374 890 383 114 24;
  • 28) 0.099 609 374 890 383 114 24 × 2 = 0 + 0.199 218 749 780 766 228 48;
  • 29) 0.199 218 749 780 766 228 48 × 2 = 0 + 0.398 437 499 561 532 456 96;
  • 30) 0.398 437 499 561 532 456 96 × 2 = 0 + 0.796 874 999 123 064 913 92;
  • 31) 0.796 874 999 123 064 913 92 × 2 = 1 + 0.593 749 998 246 129 827 84;
  • 32) 0.593 749 998 246 129 827 84 × 2 = 1 + 0.187 499 996 492 259 655 68;
  • 33) 0.187 499 996 492 259 655 68 × 2 = 0 + 0.374 999 992 984 519 311 36;
  • 34) 0.374 999 992 984 519 311 36 × 2 = 0 + 0.749 999 985 969 038 622 72;
  • 35) 0.749 999 985 969 038 622 72 × 2 = 1 + 0.499 999 971 938 077 245 44;
  • 36) 0.499 999 971 938 077 245 44 × 2 = 0 + 0.999 999 943 876 154 490 88;
  • 37) 0.999 999 943 876 154 490 88 × 2 = 1 + 0.999 999 887 752 308 981 76;
  • 38) 0.999 999 887 752 308 981 76 × 2 = 1 + 0.999 999 775 504 617 963 52;
  • 39) 0.999 999 775 504 617 963 52 × 2 = 1 + 0.999 999 551 009 235 927 04;
  • 40) 0.999 999 551 009 235 927 04 × 2 = 1 + 0.999 999 102 018 471 854 08;
  • 41) 0.999 999 102 018 471 854 08 × 2 = 1 + 0.999 998 204 036 943 708 16;
  • 42) 0.999 998 204 036 943 708 16 × 2 = 1 + 0.999 996 408 073 887 416 32;
  • 43) 0.999 996 408 073 887 416 32 × 2 = 1 + 0.999 992 816 147 774 832 64;
  • 44) 0.999 992 816 147 774 832 64 × 2 = 1 + 0.999 985 632 295 549 665 28;
  • 45) 0.999 985 632 295 549 665 28 × 2 = 1 + 0.999 971 264 591 099 330 56;
  • 46) 0.999 971 264 591 099 330 56 × 2 = 1 + 0.999 942 529 182 198 661 12;
  • 47) 0.999 942 529 182 198 661 12 × 2 = 1 + 0.999 885 058 364 397 322 24;
  • 48) 0.999 885 058 364 397 322 24 × 2 = 1 + 0.999 770 116 728 794 644 48;
  • 49) 0.999 770 116 728 794 644 48 × 2 = 1 + 0.999 540 233 457 589 288 96;
  • 50) 0.999 540 233 457 589 288 96 × 2 = 1 + 0.999 080 466 915 178 577 92;
  • 51) 0.999 080 466 915 178 577 92 × 2 = 1 + 0.998 160 933 830 357 155 84;
  • 52) 0.998 160 933 830 357 155 84 × 2 = 1 + 0.996 321 867 660 714 311 68;
  • 53) 0.996 321 867 660 714 311 68 × 2 = 1 + 0.992 643 735 321 428 623 36;
  • 54) 0.992 643 735 321 428 623 36 × 2 = 1 + 0.985 287 470 642 857 246 72;

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:


0.000 000 000 742 147 675 83(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 83(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:


0.000 000 000 742 147 675 83(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 -0.000 000 000 742 147 675 83 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