-0.000 000 000 742 147 676 646 718 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 718 2(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
-0.000 000 000 742 147 676 646 718 2(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:

|-0.000 000 000 742 147 676 646 718 2| = 0.000 000 000 742 147 676 646 718 2


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: 0.000 000 000 742 147 676 646 718 2.

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) 0.000 000 000 742 147 676 646 718 2 × 2 = 0 + 0.000 000 001 484 295 353 293 436 4;
  • 2) 0.000 000 001 484 295 353 293 436 4 × 2 = 0 + 0.000 000 002 968 590 706 586 872 8;
  • 3) 0.000 000 002 968 590 706 586 872 8 × 2 = 0 + 0.000 000 005 937 181 413 173 745 6;
  • 4) 0.000 000 005 937 181 413 173 745 6 × 2 = 0 + 0.000 000 011 874 362 826 347 491 2;
  • 5) 0.000 000 011 874 362 826 347 491 2 × 2 = 0 + 0.000 000 023 748 725 652 694 982 4;
  • 6) 0.000 000 023 748 725 652 694 982 4 × 2 = 0 + 0.000 000 047 497 451 305 389 964 8;
  • 7) 0.000 000 047 497 451 305 389 964 8 × 2 = 0 + 0.000 000 094 994 902 610 779 929 6;
  • 8) 0.000 000 094 994 902 610 779 929 6 × 2 = 0 + 0.000 000 189 989 805 221 559 859 2;
  • 9) 0.000 000 189 989 805 221 559 859 2 × 2 = 0 + 0.000 000 379 979 610 443 119 718 4;
  • 10) 0.000 000 379 979 610 443 119 718 4 × 2 = 0 + 0.000 000 759 959 220 886 239 436 8;
  • 11) 0.000 000 759 959 220 886 239 436 8 × 2 = 0 + 0.000 001 519 918 441 772 478 873 6;
  • 12) 0.000 001 519 918 441 772 478 873 6 × 2 = 0 + 0.000 003 039 836 883 544 957 747 2;
  • 13) 0.000 003 039 836 883 544 957 747 2 × 2 = 0 + 0.000 006 079 673 767 089 915 494 4;
  • 14) 0.000 006 079 673 767 089 915 494 4 × 2 = 0 + 0.000 012 159 347 534 179 830 988 8;
  • 15) 0.000 012 159 347 534 179 830 988 8 × 2 = 0 + 0.000 024 318 695 068 359 661 977 6;
  • 16) 0.000 024 318 695 068 359 661 977 6 × 2 = 0 + 0.000 048 637 390 136 719 323 955 2;
  • 17) 0.000 048 637 390 136 719 323 955 2 × 2 = 0 + 0.000 097 274 780 273 438 647 910 4;
  • 18) 0.000 097 274 780 273 438 647 910 4 × 2 = 0 + 0.000 194 549 560 546 877 295 820 8;
  • 19) 0.000 194 549 560 546 877 295 820 8 × 2 = 0 + 0.000 389 099 121 093 754 591 641 6;
  • 20) 0.000 389 099 121 093 754 591 641 6 × 2 = 0 + 0.000 778 198 242 187 509 183 283 2;
  • 21) 0.000 778 198 242 187 509 183 283 2 × 2 = 0 + 0.001 556 396 484 375 018 366 566 4;
  • 22) 0.001 556 396 484 375 018 366 566 4 × 2 = 0 + 0.003 112 792 968 750 036 733 132 8;
  • 23) 0.003 112 792 968 750 036 733 132 8 × 2 = 0 + 0.006 225 585 937 500 073 466 265 6;
  • 24) 0.006 225 585 937 500 073 466 265 6 × 2 = 0 + 0.012 451 171 875 000 146 932 531 2;
  • 25) 0.012 451 171 875 000 146 932 531 2 × 2 = 0 + 0.024 902 343 750 000 293 865 062 4;
  • 26) 0.024 902 343 750 000 293 865 062 4 × 2 = 0 + 0.049 804 687 500 000 587 730 124 8;
  • 27) 0.049 804 687 500 000 587 730 124 8 × 2 = 0 + 0.099 609 375 000 001 175 460 249 6;
  • 28) 0.099 609 375 000 001 175 460 249 6 × 2 = 0 + 0.199 218 750 000 002 350 920 499 2;
  • 29) 0.199 218 750 000 002 350 920 499 2 × 2 = 0 + 0.398 437 500 000 004 701 840 998 4;
  • 30) 0.398 437 500 000 004 701 840 998 4 × 2 = 0 + 0.796 875 000 000 009 403 681 996 8;
  • 31) 0.796 875 000 000 009 403 681 996 8 × 2 = 1 + 0.593 750 000 000 018 807 363 993 6;
  • 32) 0.593 750 000 000 018 807 363 993 6 × 2 = 1 + 0.187 500 000 000 037 614 727 987 2;
  • 33) 0.187 500 000 000 037 614 727 987 2 × 2 = 0 + 0.375 000 000 000 075 229 455 974 4;
  • 34) 0.375 000 000 000 075 229 455 974 4 × 2 = 0 + 0.750 000 000 000 150 458 911 948 8;
  • 35) 0.750 000 000 000 150 458 911 948 8 × 2 = 1 + 0.500 000 000 000 300 917 823 897 6;
  • 36) 0.500 000 000 000 300 917 823 897 6 × 2 = 1 + 0.000 000 000 000 601 835 647 795 2;
  • 37) 0.000 000 000 000 601 835 647 795 2 × 2 = 0 + 0.000 000 000 001 203 671 295 590 4;
  • 38) 0.000 000 000 001 203 671 295 590 4 × 2 = 0 + 0.000 000 000 002 407 342 591 180 8;
  • 39) 0.000 000 000 002 407 342 591 180 8 × 2 = 0 + 0.000 000 000 004 814 685 182 361 6;
  • 40) 0.000 000 000 004 814 685 182 361 6 × 2 = 0 + 0.000 000 000 009 629 370 364 723 2;
  • 41) 0.000 000 000 009 629 370 364 723 2 × 2 = 0 + 0.000 000 000 019 258 740 729 446 4;
  • 42) 0.000 000 000 019 258 740 729 446 4 × 2 = 0 + 0.000 000 000 038 517 481 458 892 8;
  • 43) 0.000 000 000 038 517 481 458 892 8 × 2 = 0 + 0.000 000 000 077 034 962 917 785 6;
  • 44) 0.000 000 000 077 034 962 917 785 6 × 2 = 0 + 0.000 000 000 154 069 925 835 571 2;
  • 45) 0.000 000 000 154 069 925 835 571 2 × 2 = 0 + 0.000 000 000 308 139 851 671 142 4;
  • 46) 0.000 000 000 308 139 851 671 142 4 × 2 = 0 + 0.000 000 000 616 279 703 342 284 8;
  • 47) 0.000 000 000 616 279 703 342 284 8 × 2 = 0 + 0.000 000 001 232 559 406 684 569 6;
  • 48) 0.000 000 001 232 559 406 684 569 6 × 2 = 0 + 0.000 000 002 465 118 813 369 139 2;
  • 49) 0.000 000 002 465 118 813 369 139 2 × 2 = 0 + 0.000 000 004 930 237 626 738 278 4;
  • 50) 0.000 000 004 930 237 626 738 278 4 × 2 = 0 + 0.000 000 009 860 475 253 476 556 8;
  • 51) 0.000 000 009 860 475 253 476 556 8 × 2 = 0 + 0.000 000 019 720 950 506 953 113 6;
  • 52) 0.000 000 019 720 950 506 953 113 6 × 2 = 0 + 0.000 000 039 441 901 013 906 227 2;
  • 53) 0.000 000 039 441 901 013 906 227 2 × 2 = 0 + 0.000 000 078 883 802 027 812 454 4;
  • 54) 0.000 000 078 883 802 027 812 454 4 × 2 = 0 + 0.000 000 157 767 604 055 624 908 8;

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:


0.000 000 000 742 147 676 646 718 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 718 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 718 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 718 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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