-0.000 000 000 742 147 675 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 79(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 79(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 79| = 0.000 000 000 742 147 675 79


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 79.

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 79 × 2 = 0 + 0.000 000 001 484 295 351 58;
  • 2) 0.000 000 001 484 295 351 58 × 2 = 0 + 0.000 000 002 968 590 703 16;
  • 3) 0.000 000 002 968 590 703 16 × 2 = 0 + 0.000 000 005 937 181 406 32;
  • 4) 0.000 000 005 937 181 406 32 × 2 = 0 + 0.000 000 011 874 362 812 64;
  • 5) 0.000 000 011 874 362 812 64 × 2 = 0 + 0.000 000 023 748 725 625 28;
  • 6) 0.000 000 023 748 725 625 28 × 2 = 0 + 0.000 000 047 497 451 250 56;
  • 7) 0.000 000 047 497 451 250 56 × 2 = 0 + 0.000 000 094 994 902 501 12;
  • 8) 0.000 000 094 994 902 501 12 × 2 = 0 + 0.000 000 189 989 805 002 24;
  • 9) 0.000 000 189 989 805 002 24 × 2 = 0 + 0.000 000 379 979 610 004 48;
  • 10) 0.000 000 379 979 610 004 48 × 2 = 0 + 0.000 000 759 959 220 008 96;
  • 11) 0.000 000 759 959 220 008 96 × 2 = 0 + 0.000 001 519 918 440 017 92;
  • 12) 0.000 001 519 918 440 017 92 × 2 = 0 + 0.000 003 039 836 880 035 84;
  • 13) 0.000 003 039 836 880 035 84 × 2 = 0 + 0.000 006 079 673 760 071 68;
  • 14) 0.000 006 079 673 760 071 68 × 2 = 0 + 0.000 012 159 347 520 143 36;
  • 15) 0.000 012 159 347 520 143 36 × 2 = 0 + 0.000 024 318 695 040 286 72;
  • 16) 0.000 024 318 695 040 286 72 × 2 = 0 + 0.000 048 637 390 080 573 44;
  • 17) 0.000 048 637 390 080 573 44 × 2 = 0 + 0.000 097 274 780 161 146 88;
  • 18) 0.000 097 274 780 161 146 88 × 2 = 0 + 0.000 194 549 560 322 293 76;
  • 19) 0.000 194 549 560 322 293 76 × 2 = 0 + 0.000 389 099 120 644 587 52;
  • 20) 0.000 389 099 120 644 587 52 × 2 = 0 + 0.000 778 198 241 289 175 04;
  • 21) 0.000 778 198 241 289 175 04 × 2 = 0 + 0.001 556 396 482 578 350 08;
  • 22) 0.001 556 396 482 578 350 08 × 2 = 0 + 0.003 112 792 965 156 700 16;
  • 23) 0.003 112 792 965 156 700 16 × 2 = 0 + 0.006 225 585 930 313 400 32;
  • 24) 0.006 225 585 930 313 400 32 × 2 = 0 + 0.012 451 171 860 626 800 64;
  • 25) 0.012 451 171 860 626 800 64 × 2 = 0 + 0.024 902 343 721 253 601 28;
  • 26) 0.024 902 343 721 253 601 28 × 2 = 0 + 0.049 804 687 442 507 202 56;
  • 27) 0.049 804 687 442 507 202 56 × 2 = 0 + 0.099 609 374 885 014 405 12;
  • 28) 0.099 609 374 885 014 405 12 × 2 = 0 + 0.199 218 749 770 028 810 24;
  • 29) 0.199 218 749 770 028 810 24 × 2 = 0 + 0.398 437 499 540 057 620 48;
  • 30) 0.398 437 499 540 057 620 48 × 2 = 0 + 0.796 874 999 080 115 240 96;
  • 31) 0.796 874 999 080 115 240 96 × 2 = 1 + 0.593 749 998 160 230 481 92;
  • 32) 0.593 749 998 160 230 481 92 × 2 = 1 + 0.187 499 996 320 460 963 84;
  • 33) 0.187 499 996 320 460 963 84 × 2 = 0 + 0.374 999 992 640 921 927 68;
  • 34) 0.374 999 992 640 921 927 68 × 2 = 0 + 0.749 999 985 281 843 855 36;
  • 35) 0.749 999 985 281 843 855 36 × 2 = 1 + 0.499 999 970 563 687 710 72;
  • 36) 0.499 999 970 563 687 710 72 × 2 = 0 + 0.999 999 941 127 375 421 44;
  • 37) 0.999 999 941 127 375 421 44 × 2 = 1 + 0.999 999 882 254 750 842 88;
  • 38) 0.999 999 882 254 750 842 88 × 2 = 1 + 0.999 999 764 509 501 685 76;
  • 39) 0.999 999 764 509 501 685 76 × 2 = 1 + 0.999 999 529 019 003 371 52;
  • 40) 0.999 999 529 019 003 371 52 × 2 = 1 + 0.999 999 058 038 006 743 04;
  • 41) 0.999 999 058 038 006 743 04 × 2 = 1 + 0.999 998 116 076 013 486 08;
  • 42) 0.999 998 116 076 013 486 08 × 2 = 1 + 0.999 996 232 152 026 972 16;
  • 43) 0.999 996 232 152 026 972 16 × 2 = 1 + 0.999 992 464 304 053 944 32;
  • 44) 0.999 992 464 304 053 944 32 × 2 = 1 + 0.999 984 928 608 107 888 64;
  • 45) 0.999 984 928 608 107 888 64 × 2 = 1 + 0.999 969 857 216 215 777 28;
  • 46) 0.999 969 857 216 215 777 28 × 2 = 1 + 0.999 939 714 432 431 554 56;
  • 47) 0.999 939 714 432 431 554 56 × 2 = 1 + 0.999 879 428 864 863 109 12;
  • 48) 0.999 879 428 864 863 109 12 × 2 = 1 + 0.999 758 857 729 726 218 24;
  • 49) 0.999 758 857 729 726 218 24 × 2 = 1 + 0.999 517 715 459 452 436 48;
  • 50) 0.999 517 715 459 452 436 48 × 2 = 1 + 0.999 035 430 918 904 872 96;
  • 51) 0.999 035 430 918 904 872 96 × 2 = 1 + 0.998 070 861 837 809 745 92;
  • 52) 0.998 070 861 837 809 745 92 × 2 = 1 + 0.996 141 723 675 619 491 84;
  • 53) 0.996 141 723 675 619 491 84 × 2 = 1 + 0.992 283 447 351 238 983 68;
  • 54) 0.992 283 447 351 238 983 68 × 2 = 1 + 0.984 566 894 702 477 967 36;

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 79(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 79(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 79(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 79 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