-0.000 000 000 742 147 676 646 723 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 723 6(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
-0.000 000 000 742 147 676 646 723 6(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:

|-0.000 000 000 742 147 676 646 723 6| = 0.000 000 000 742 147 676 646 723 6


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: 0.000 000 000 742 147 676 646 723 6.

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) 0.000 000 000 742 147 676 646 723 6 × 2 = 0 + 0.000 000 001 484 295 353 293 447 2;
  • 2) 0.000 000 001 484 295 353 293 447 2 × 2 = 0 + 0.000 000 002 968 590 706 586 894 4;
  • 3) 0.000 000 002 968 590 706 586 894 4 × 2 = 0 + 0.000 000 005 937 181 413 173 788 8;
  • 4) 0.000 000 005 937 181 413 173 788 8 × 2 = 0 + 0.000 000 011 874 362 826 347 577 6;
  • 5) 0.000 000 011 874 362 826 347 577 6 × 2 = 0 + 0.000 000 023 748 725 652 695 155 2;
  • 6) 0.000 000 023 748 725 652 695 155 2 × 2 = 0 + 0.000 000 047 497 451 305 390 310 4;
  • 7) 0.000 000 047 497 451 305 390 310 4 × 2 = 0 + 0.000 000 094 994 902 610 780 620 8;
  • 8) 0.000 000 094 994 902 610 780 620 8 × 2 = 0 + 0.000 000 189 989 805 221 561 241 6;
  • 9) 0.000 000 189 989 805 221 561 241 6 × 2 = 0 + 0.000 000 379 979 610 443 122 483 2;
  • 10) 0.000 000 379 979 610 443 122 483 2 × 2 = 0 + 0.000 000 759 959 220 886 244 966 4;
  • 11) 0.000 000 759 959 220 886 244 966 4 × 2 = 0 + 0.000 001 519 918 441 772 489 932 8;
  • 12) 0.000 001 519 918 441 772 489 932 8 × 2 = 0 + 0.000 003 039 836 883 544 979 865 6;
  • 13) 0.000 003 039 836 883 544 979 865 6 × 2 = 0 + 0.000 006 079 673 767 089 959 731 2;
  • 14) 0.000 006 079 673 767 089 959 731 2 × 2 = 0 + 0.000 012 159 347 534 179 919 462 4;
  • 15) 0.000 012 159 347 534 179 919 462 4 × 2 = 0 + 0.000 024 318 695 068 359 838 924 8;
  • 16) 0.000 024 318 695 068 359 838 924 8 × 2 = 0 + 0.000 048 637 390 136 719 677 849 6;
  • 17) 0.000 048 637 390 136 719 677 849 6 × 2 = 0 + 0.000 097 274 780 273 439 355 699 2;
  • 18) 0.000 097 274 780 273 439 355 699 2 × 2 = 0 + 0.000 194 549 560 546 878 711 398 4;
  • 19) 0.000 194 549 560 546 878 711 398 4 × 2 = 0 + 0.000 389 099 121 093 757 422 796 8;
  • 20) 0.000 389 099 121 093 757 422 796 8 × 2 = 0 + 0.000 778 198 242 187 514 845 593 6;
  • 21) 0.000 778 198 242 187 514 845 593 6 × 2 = 0 + 0.001 556 396 484 375 029 691 187 2;
  • 22) 0.001 556 396 484 375 029 691 187 2 × 2 = 0 + 0.003 112 792 968 750 059 382 374 4;
  • 23) 0.003 112 792 968 750 059 382 374 4 × 2 = 0 + 0.006 225 585 937 500 118 764 748 8;
  • 24) 0.006 225 585 937 500 118 764 748 8 × 2 = 0 + 0.012 451 171 875 000 237 529 497 6;
  • 25) 0.012 451 171 875 000 237 529 497 6 × 2 = 0 + 0.024 902 343 750 000 475 058 995 2;
  • 26) 0.024 902 343 750 000 475 058 995 2 × 2 = 0 + 0.049 804 687 500 000 950 117 990 4;
  • 27) 0.049 804 687 500 000 950 117 990 4 × 2 = 0 + 0.099 609 375 000 001 900 235 980 8;
  • 28) 0.099 609 375 000 001 900 235 980 8 × 2 = 0 + 0.199 218 750 000 003 800 471 961 6;
  • 29) 0.199 218 750 000 003 800 471 961 6 × 2 = 0 + 0.398 437 500 000 007 600 943 923 2;
  • 30) 0.398 437 500 000 007 600 943 923 2 × 2 = 0 + 0.796 875 000 000 015 201 887 846 4;
  • 31) 0.796 875 000 000 015 201 887 846 4 × 2 = 1 + 0.593 750 000 000 030 403 775 692 8;
  • 32) 0.593 750 000 000 030 403 775 692 8 × 2 = 1 + 0.187 500 000 000 060 807 551 385 6;
  • 33) 0.187 500 000 000 060 807 551 385 6 × 2 = 0 + 0.375 000 000 000 121 615 102 771 2;
  • 34) 0.375 000 000 000 121 615 102 771 2 × 2 = 0 + 0.750 000 000 000 243 230 205 542 4;
  • 35) 0.750 000 000 000 243 230 205 542 4 × 2 = 1 + 0.500 000 000 000 486 460 411 084 8;
  • 36) 0.500 000 000 000 486 460 411 084 8 × 2 = 1 + 0.000 000 000 000 972 920 822 169 6;
  • 37) 0.000 000 000 000 972 920 822 169 6 × 2 = 0 + 0.000 000 000 001 945 841 644 339 2;
  • 38) 0.000 000 000 001 945 841 644 339 2 × 2 = 0 + 0.000 000 000 003 891 683 288 678 4;
  • 39) 0.000 000 000 003 891 683 288 678 4 × 2 = 0 + 0.000 000 000 007 783 366 577 356 8;
  • 40) 0.000 000 000 007 783 366 577 356 8 × 2 = 0 + 0.000 000 000 015 566 733 154 713 6;
  • 41) 0.000 000 000 015 566 733 154 713 6 × 2 = 0 + 0.000 000 000 031 133 466 309 427 2;
  • 42) 0.000 000 000 031 133 466 309 427 2 × 2 = 0 + 0.000 000 000 062 266 932 618 854 4;
  • 43) 0.000 000 000 062 266 932 618 854 4 × 2 = 0 + 0.000 000 000 124 533 865 237 708 8;
  • 44) 0.000 000 000 124 533 865 237 708 8 × 2 = 0 + 0.000 000 000 249 067 730 475 417 6;
  • 45) 0.000 000 000 249 067 730 475 417 6 × 2 = 0 + 0.000 000 000 498 135 460 950 835 2;
  • 46) 0.000 000 000 498 135 460 950 835 2 × 2 = 0 + 0.000 000 000 996 270 921 901 670 4;
  • 47) 0.000 000 000 996 270 921 901 670 4 × 2 = 0 + 0.000 000 001 992 541 843 803 340 8;
  • 48) 0.000 000 001 992 541 843 803 340 8 × 2 = 0 + 0.000 000 003 985 083 687 606 681 6;
  • 49) 0.000 000 003 985 083 687 606 681 6 × 2 = 0 + 0.000 000 007 970 167 375 213 363 2;
  • 50) 0.000 000 007 970 167 375 213 363 2 × 2 = 0 + 0.000 000 015 940 334 750 426 726 4;
  • 51) 0.000 000 015 940 334 750 426 726 4 × 2 = 0 + 0.000 000 031 880 669 500 853 452 8;
  • 52) 0.000 000 031 880 669 500 853 452 8 × 2 = 0 + 0.000 000 063 761 339 001 706 905 6;
  • 53) 0.000 000 063 761 339 001 706 905 6 × 2 = 0 + 0.000 000 127 522 678 003 413 811 2;
  • 54) 0.000 000 127 522 678 003 413 811 2 × 2 = 0 + 0.000 000 255 045 356 006 827 622 4;

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:


0.000 000 000 742 147 676 646 723 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 723 6(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:


0.000 000 000 742 147 676 646 723 6(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 -0.000 000 000 742 147 676 646 723 6 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