-0.000 000 000 742 147 676 646 708 94 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 708 94(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
-0.000 000 000 742 147 676 646 708 94(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:

|-0.000 000 000 742 147 676 646 708 94| = 0.000 000 000 742 147 676 646 708 94


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: 0.000 000 000 742 147 676 646 708 94.

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) 0.000 000 000 742 147 676 646 708 94 × 2 = 0 + 0.000 000 001 484 295 353 293 417 88;
  • 2) 0.000 000 001 484 295 353 293 417 88 × 2 = 0 + 0.000 000 002 968 590 706 586 835 76;
  • 3) 0.000 000 002 968 590 706 586 835 76 × 2 = 0 + 0.000 000 005 937 181 413 173 671 52;
  • 4) 0.000 000 005 937 181 413 173 671 52 × 2 = 0 + 0.000 000 011 874 362 826 347 343 04;
  • 5) 0.000 000 011 874 362 826 347 343 04 × 2 = 0 + 0.000 000 023 748 725 652 694 686 08;
  • 6) 0.000 000 023 748 725 652 694 686 08 × 2 = 0 + 0.000 000 047 497 451 305 389 372 16;
  • 7) 0.000 000 047 497 451 305 389 372 16 × 2 = 0 + 0.000 000 094 994 902 610 778 744 32;
  • 8) 0.000 000 094 994 902 610 778 744 32 × 2 = 0 + 0.000 000 189 989 805 221 557 488 64;
  • 9) 0.000 000 189 989 805 221 557 488 64 × 2 = 0 + 0.000 000 379 979 610 443 114 977 28;
  • 10) 0.000 000 379 979 610 443 114 977 28 × 2 = 0 + 0.000 000 759 959 220 886 229 954 56;
  • 11) 0.000 000 759 959 220 886 229 954 56 × 2 = 0 + 0.000 001 519 918 441 772 459 909 12;
  • 12) 0.000 001 519 918 441 772 459 909 12 × 2 = 0 + 0.000 003 039 836 883 544 919 818 24;
  • 13) 0.000 003 039 836 883 544 919 818 24 × 2 = 0 + 0.000 006 079 673 767 089 839 636 48;
  • 14) 0.000 006 079 673 767 089 839 636 48 × 2 = 0 + 0.000 012 159 347 534 179 679 272 96;
  • 15) 0.000 012 159 347 534 179 679 272 96 × 2 = 0 + 0.000 024 318 695 068 359 358 545 92;
  • 16) 0.000 024 318 695 068 359 358 545 92 × 2 = 0 + 0.000 048 637 390 136 718 717 091 84;
  • 17) 0.000 048 637 390 136 718 717 091 84 × 2 = 0 + 0.000 097 274 780 273 437 434 183 68;
  • 18) 0.000 097 274 780 273 437 434 183 68 × 2 = 0 + 0.000 194 549 560 546 874 868 367 36;
  • 19) 0.000 194 549 560 546 874 868 367 36 × 2 = 0 + 0.000 389 099 121 093 749 736 734 72;
  • 20) 0.000 389 099 121 093 749 736 734 72 × 2 = 0 + 0.000 778 198 242 187 499 473 469 44;
  • 21) 0.000 778 198 242 187 499 473 469 44 × 2 = 0 + 0.001 556 396 484 374 998 946 938 88;
  • 22) 0.001 556 396 484 374 998 946 938 88 × 2 = 0 + 0.003 112 792 968 749 997 893 877 76;
  • 23) 0.003 112 792 968 749 997 893 877 76 × 2 = 0 + 0.006 225 585 937 499 995 787 755 52;
  • 24) 0.006 225 585 937 499 995 787 755 52 × 2 = 0 + 0.012 451 171 874 999 991 575 511 04;
  • 25) 0.012 451 171 874 999 991 575 511 04 × 2 = 0 + 0.024 902 343 749 999 983 151 022 08;
  • 26) 0.024 902 343 749 999 983 151 022 08 × 2 = 0 + 0.049 804 687 499 999 966 302 044 16;
  • 27) 0.049 804 687 499 999 966 302 044 16 × 2 = 0 + 0.099 609 374 999 999 932 604 088 32;
  • 28) 0.099 609 374 999 999 932 604 088 32 × 2 = 0 + 0.199 218 749 999 999 865 208 176 64;
  • 29) 0.199 218 749 999 999 865 208 176 64 × 2 = 0 + 0.398 437 499 999 999 730 416 353 28;
  • 30) 0.398 437 499 999 999 730 416 353 28 × 2 = 0 + 0.796 874 999 999 999 460 832 706 56;
  • 31) 0.796 874 999 999 999 460 832 706 56 × 2 = 1 + 0.593 749 999 999 998 921 665 413 12;
  • 32) 0.593 749 999 999 998 921 665 413 12 × 2 = 1 + 0.187 499 999 999 997 843 330 826 24;
  • 33) 0.187 499 999 999 997 843 330 826 24 × 2 = 0 + 0.374 999 999 999 995 686 661 652 48;
  • 34) 0.374 999 999 999 995 686 661 652 48 × 2 = 0 + 0.749 999 999 999 991 373 323 304 96;
  • 35) 0.749 999 999 999 991 373 323 304 96 × 2 = 1 + 0.499 999 999 999 982 746 646 609 92;
  • 36) 0.499 999 999 999 982 746 646 609 92 × 2 = 0 + 0.999 999 999 999 965 493 293 219 84;
  • 37) 0.999 999 999 999 965 493 293 219 84 × 2 = 1 + 0.999 999 999 999 930 986 586 439 68;
  • 38) 0.999 999 999 999 930 986 586 439 68 × 2 = 1 + 0.999 999 999 999 861 973 172 879 36;
  • 39) 0.999 999 999 999 861 973 172 879 36 × 2 = 1 + 0.999 999 999 999 723 946 345 758 72;
  • 40) 0.999 999 999 999 723 946 345 758 72 × 2 = 1 + 0.999 999 999 999 447 892 691 517 44;
  • 41) 0.999 999 999 999 447 892 691 517 44 × 2 = 1 + 0.999 999 999 998 895 785 383 034 88;
  • 42) 0.999 999 999 998 895 785 383 034 88 × 2 = 1 + 0.999 999 999 997 791 570 766 069 76;
  • 43) 0.999 999 999 997 791 570 766 069 76 × 2 = 1 + 0.999 999 999 995 583 141 532 139 52;
  • 44) 0.999 999 999 995 583 141 532 139 52 × 2 = 1 + 0.999 999 999 991 166 283 064 279 04;
  • 45) 0.999 999 999 991 166 283 064 279 04 × 2 = 1 + 0.999 999 999 982 332 566 128 558 08;
  • 46) 0.999 999 999 982 332 566 128 558 08 × 2 = 1 + 0.999 999 999 964 665 132 257 116 16;
  • 47) 0.999 999 999 964 665 132 257 116 16 × 2 = 1 + 0.999 999 999 929 330 264 514 232 32;
  • 48) 0.999 999 999 929 330 264 514 232 32 × 2 = 1 + 0.999 999 999 858 660 529 028 464 64;
  • 49) 0.999 999 999 858 660 529 028 464 64 × 2 = 1 + 0.999 999 999 717 321 058 056 929 28;
  • 50) 0.999 999 999 717 321 058 056 929 28 × 2 = 1 + 0.999 999 999 434 642 116 113 858 56;
  • 51) 0.999 999 999 434 642 116 113 858 56 × 2 = 1 + 0.999 999 998 869 284 232 227 717 12;
  • 52) 0.999 999 998 869 284 232 227 717 12 × 2 = 1 + 0.999 999 997 738 568 464 455 434 24;
  • 53) 0.999 999 997 738 568 464 455 434 24 × 2 = 1 + 0.999 999 995 477 136 928 910 868 48;
  • 54) 0.999 999 995 477 136 928 910 868 48 × 2 = 1 + 0.999 999 990 954 273 857 821 736 96;

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:


0.000 000 000 742 147 676 646 708 94(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 708 94(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:


0.000 000 000 742 147 676 646 708 94(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 -0.000 000 000 742 147 676 646 708 94 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