-0.000 000 000 742 147 676 628 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 628 8(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
-0.000 000 000 742 147 676 628 8(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:

|-0.000 000 000 742 147 676 628 8| = 0.000 000 000 742 147 676 628 8


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: 0.000 000 000 742 147 676 628 8.

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) 0.000 000 000 742 147 676 628 8 × 2 = 0 + 0.000 000 001 484 295 353 257 6;
  • 2) 0.000 000 001 484 295 353 257 6 × 2 = 0 + 0.000 000 002 968 590 706 515 2;
  • 3) 0.000 000 002 968 590 706 515 2 × 2 = 0 + 0.000 000 005 937 181 413 030 4;
  • 4) 0.000 000 005 937 181 413 030 4 × 2 = 0 + 0.000 000 011 874 362 826 060 8;
  • 5) 0.000 000 011 874 362 826 060 8 × 2 = 0 + 0.000 000 023 748 725 652 121 6;
  • 6) 0.000 000 023 748 725 652 121 6 × 2 = 0 + 0.000 000 047 497 451 304 243 2;
  • 7) 0.000 000 047 497 451 304 243 2 × 2 = 0 + 0.000 000 094 994 902 608 486 4;
  • 8) 0.000 000 094 994 902 608 486 4 × 2 = 0 + 0.000 000 189 989 805 216 972 8;
  • 9) 0.000 000 189 989 805 216 972 8 × 2 = 0 + 0.000 000 379 979 610 433 945 6;
  • 10) 0.000 000 379 979 610 433 945 6 × 2 = 0 + 0.000 000 759 959 220 867 891 2;
  • 11) 0.000 000 759 959 220 867 891 2 × 2 = 0 + 0.000 001 519 918 441 735 782 4;
  • 12) 0.000 001 519 918 441 735 782 4 × 2 = 0 + 0.000 003 039 836 883 471 564 8;
  • 13) 0.000 003 039 836 883 471 564 8 × 2 = 0 + 0.000 006 079 673 766 943 129 6;
  • 14) 0.000 006 079 673 766 943 129 6 × 2 = 0 + 0.000 012 159 347 533 886 259 2;
  • 15) 0.000 012 159 347 533 886 259 2 × 2 = 0 + 0.000 024 318 695 067 772 518 4;
  • 16) 0.000 024 318 695 067 772 518 4 × 2 = 0 + 0.000 048 637 390 135 545 036 8;
  • 17) 0.000 048 637 390 135 545 036 8 × 2 = 0 + 0.000 097 274 780 271 090 073 6;
  • 18) 0.000 097 274 780 271 090 073 6 × 2 = 0 + 0.000 194 549 560 542 180 147 2;
  • 19) 0.000 194 549 560 542 180 147 2 × 2 = 0 + 0.000 389 099 121 084 360 294 4;
  • 20) 0.000 389 099 121 084 360 294 4 × 2 = 0 + 0.000 778 198 242 168 720 588 8;
  • 21) 0.000 778 198 242 168 720 588 8 × 2 = 0 + 0.001 556 396 484 337 441 177 6;
  • 22) 0.001 556 396 484 337 441 177 6 × 2 = 0 + 0.003 112 792 968 674 882 355 2;
  • 23) 0.003 112 792 968 674 882 355 2 × 2 = 0 + 0.006 225 585 937 349 764 710 4;
  • 24) 0.006 225 585 937 349 764 710 4 × 2 = 0 + 0.012 451 171 874 699 529 420 8;
  • 25) 0.012 451 171 874 699 529 420 8 × 2 = 0 + 0.024 902 343 749 399 058 841 6;
  • 26) 0.024 902 343 749 399 058 841 6 × 2 = 0 + 0.049 804 687 498 798 117 683 2;
  • 27) 0.049 804 687 498 798 117 683 2 × 2 = 0 + 0.099 609 374 997 596 235 366 4;
  • 28) 0.099 609 374 997 596 235 366 4 × 2 = 0 + 0.199 218 749 995 192 470 732 8;
  • 29) 0.199 218 749 995 192 470 732 8 × 2 = 0 + 0.398 437 499 990 384 941 465 6;
  • 30) 0.398 437 499 990 384 941 465 6 × 2 = 0 + 0.796 874 999 980 769 882 931 2;
  • 31) 0.796 874 999 980 769 882 931 2 × 2 = 1 + 0.593 749 999 961 539 765 862 4;
  • 32) 0.593 749 999 961 539 765 862 4 × 2 = 1 + 0.187 499 999 923 079 531 724 8;
  • 33) 0.187 499 999 923 079 531 724 8 × 2 = 0 + 0.374 999 999 846 159 063 449 6;
  • 34) 0.374 999 999 846 159 063 449 6 × 2 = 0 + 0.749 999 999 692 318 126 899 2;
  • 35) 0.749 999 999 692 318 126 899 2 × 2 = 1 + 0.499 999 999 384 636 253 798 4;
  • 36) 0.499 999 999 384 636 253 798 4 × 2 = 0 + 0.999 999 998 769 272 507 596 8;
  • 37) 0.999 999 998 769 272 507 596 8 × 2 = 1 + 0.999 999 997 538 545 015 193 6;
  • 38) 0.999 999 997 538 545 015 193 6 × 2 = 1 + 0.999 999 995 077 090 030 387 2;
  • 39) 0.999 999 995 077 090 030 387 2 × 2 = 1 + 0.999 999 990 154 180 060 774 4;
  • 40) 0.999 999 990 154 180 060 774 4 × 2 = 1 + 0.999 999 980 308 360 121 548 8;
  • 41) 0.999 999 980 308 360 121 548 8 × 2 = 1 + 0.999 999 960 616 720 243 097 6;
  • 42) 0.999 999 960 616 720 243 097 6 × 2 = 1 + 0.999 999 921 233 440 486 195 2;
  • 43) 0.999 999 921 233 440 486 195 2 × 2 = 1 + 0.999 999 842 466 880 972 390 4;
  • 44) 0.999 999 842 466 880 972 390 4 × 2 = 1 + 0.999 999 684 933 761 944 780 8;
  • 45) 0.999 999 684 933 761 944 780 8 × 2 = 1 + 0.999 999 369 867 523 889 561 6;
  • 46) 0.999 999 369 867 523 889 561 6 × 2 = 1 + 0.999 998 739 735 047 779 123 2;
  • 47) 0.999 998 739 735 047 779 123 2 × 2 = 1 + 0.999 997 479 470 095 558 246 4;
  • 48) 0.999 997 479 470 095 558 246 4 × 2 = 1 + 0.999 994 958 940 191 116 492 8;
  • 49) 0.999 994 958 940 191 116 492 8 × 2 = 1 + 0.999 989 917 880 382 232 985 6;
  • 50) 0.999 989 917 880 382 232 985 6 × 2 = 1 + 0.999 979 835 760 764 465 971 2;
  • 51) 0.999 979 835 760 764 465 971 2 × 2 = 1 + 0.999 959 671 521 528 931 942 4;
  • 52) 0.999 959 671 521 528 931 942 4 × 2 = 1 + 0.999 919 343 043 057 863 884 8;
  • 53) 0.999 919 343 043 057 863 884 8 × 2 = 1 + 0.999 838 686 086 115 727 769 6;
  • 54) 0.999 838 686 086 115 727 769 6 × 2 = 1 + 0.999 677 372 172 231 455 539 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:


0.000 000 000 742 147 676 628 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 628 8(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:


0.000 000 000 742 147 676 628 8(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 -0.000 000 000 742 147 676 628 8 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