-0.000 000 000 742 147 676 645 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645(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
-0.000 000 000 742 147 676 645(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:

|-0.000 000 000 742 147 676 645| = 0.000 000 000 742 147 676 645


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: 0.000 000 000 742 147 676 645.

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) 0.000 000 000 742 147 676 645 × 2 = 0 + 0.000 000 001 484 295 353 29;
  • 2) 0.000 000 001 484 295 353 29 × 2 = 0 + 0.000 000 002 968 590 706 58;
  • 3) 0.000 000 002 968 590 706 58 × 2 = 0 + 0.000 000 005 937 181 413 16;
  • 4) 0.000 000 005 937 181 413 16 × 2 = 0 + 0.000 000 011 874 362 826 32;
  • 5) 0.000 000 011 874 362 826 32 × 2 = 0 + 0.000 000 023 748 725 652 64;
  • 6) 0.000 000 023 748 725 652 64 × 2 = 0 + 0.000 000 047 497 451 305 28;
  • 7) 0.000 000 047 497 451 305 28 × 2 = 0 + 0.000 000 094 994 902 610 56;
  • 8) 0.000 000 094 994 902 610 56 × 2 = 0 + 0.000 000 189 989 805 221 12;
  • 9) 0.000 000 189 989 805 221 12 × 2 = 0 + 0.000 000 379 979 610 442 24;
  • 10) 0.000 000 379 979 610 442 24 × 2 = 0 + 0.000 000 759 959 220 884 48;
  • 11) 0.000 000 759 959 220 884 48 × 2 = 0 + 0.000 001 519 918 441 768 96;
  • 12) 0.000 001 519 918 441 768 96 × 2 = 0 + 0.000 003 039 836 883 537 92;
  • 13) 0.000 003 039 836 883 537 92 × 2 = 0 + 0.000 006 079 673 767 075 84;
  • 14) 0.000 006 079 673 767 075 84 × 2 = 0 + 0.000 012 159 347 534 151 68;
  • 15) 0.000 012 159 347 534 151 68 × 2 = 0 + 0.000 024 318 695 068 303 36;
  • 16) 0.000 024 318 695 068 303 36 × 2 = 0 + 0.000 048 637 390 136 606 72;
  • 17) 0.000 048 637 390 136 606 72 × 2 = 0 + 0.000 097 274 780 273 213 44;
  • 18) 0.000 097 274 780 273 213 44 × 2 = 0 + 0.000 194 549 560 546 426 88;
  • 19) 0.000 194 549 560 546 426 88 × 2 = 0 + 0.000 389 099 121 092 853 76;
  • 20) 0.000 389 099 121 092 853 76 × 2 = 0 + 0.000 778 198 242 185 707 52;
  • 21) 0.000 778 198 242 185 707 52 × 2 = 0 + 0.001 556 396 484 371 415 04;
  • 22) 0.001 556 396 484 371 415 04 × 2 = 0 + 0.003 112 792 968 742 830 08;
  • 23) 0.003 112 792 968 742 830 08 × 2 = 0 + 0.006 225 585 937 485 660 16;
  • 24) 0.006 225 585 937 485 660 16 × 2 = 0 + 0.012 451 171 874 971 320 32;
  • 25) 0.012 451 171 874 971 320 32 × 2 = 0 + 0.024 902 343 749 942 640 64;
  • 26) 0.024 902 343 749 942 640 64 × 2 = 0 + 0.049 804 687 499 885 281 28;
  • 27) 0.049 804 687 499 885 281 28 × 2 = 0 + 0.099 609 374 999 770 562 56;
  • 28) 0.099 609 374 999 770 562 56 × 2 = 0 + 0.199 218 749 999 541 125 12;
  • 29) 0.199 218 749 999 541 125 12 × 2 = 0 + 0.398 437 499 999 082 250 24;
  • 30) 0.398 437 499 999 082 250 24 × 2 = 0 + 0.796 874 999 998 164 500 48;
  • 31) 0.796 874 999 998 164 500 48 × 2 = 1 + 0.593 749 999 996 329 000 96;
  • 32) 0.593 749 999 996 329 000 96 × 2 = 1 + 0.187 499 999 992 658 001 92;
  • 33) 0.187 499 999 992 658 001 92 × 2 = 0 + 0.374 999 999 985 316 003 84;
  • 34) 0.374 999 999 985 316 003 84 × 2 = 0 + 0.749 999 999 970 632 007 68;
  • 35) 0.749 999 999 970 632 007 68 × 2 = 1 + 0.499 999 999 941 264 015 36;
  • 36) 0.499 999 999 941 264 015 36 × 2 = 0 + 0.999 999 999 882 528 030 72;
  • 37) 0.999 999 999 882 528 030 72 × 2 = 1 + 0.999 999 999 765 056 061 44;
  • 38) 0.999 999 999 765 056 061 44 × 2 = 1 + 0.999 999 999 530 112 122 88;
  • 39) 0.999 999 999 530 112 122 88 × 2 = 1 + 0.999 999 999 060 224 245 76;
  • 40) 0.999 999 999 060 224 245 76 × 2 = 1 + 0.999 999 998 120 448 491 52;
  • 41) 0.999 999 998 120 448 491 52 × 2 = 1 + 0.999 999 996 240 896 983 04;
  • 42) 0.999 999 996 240 896 983 04 × 2 = 1 + 0.999 999 992 481 793 966 08;
  • 43) 0.999 999 992 481 793 966 08 × 2 = 1 + 0.999 999 984 963 587 932 16;
  • 44) 0.999 999 984 963 587 932 16 × 2 = 1 + 0.999 999 969 927 175 864 32;
  • 45) 0.999 999 969 927 175 864 32 × 2 = 1 + 0.999 999 939 854 351 728 64;
  • 46) 0.999 999 939 854 351 728 64 × 2 = 1 + 0.999 999 879 708 703 457 28;
  • 47) 0.999 999 879 708 703 457 28 × 2 = 1 + 0.999 999 759 417 406 914 56;
  • 48) 0.999 999 759 417 406 914 56 × 2 = 1 + 0.999 999 518 834 813 829 12;
  • 49) 0.999 999 518 834 813 829 12 × 2 = 1 + 0.999 999 037 669 627 658 24;
  • 50) 0.999 999 037 669 627 658 24 × 2 = 1 + 0.999 998 075 339 255 316 48;
  • 51) 0.999 998 075 339 255 316 48 × 2 = 1 + 0.999 996 150 678 510 632 96;
  • 52) 0.999 996 150 678 510 632 96 × 2 = 1 + 0.999 992 301 357 021 265 92;
  • 53) 0.999 992 301 357 021 265 92 × 2 = 1 + 0.999 984 602 714 042 531 84;
  • 54) 0.999 984 602 714 042 531 84 × 2 = 1 + 0.999 969 205 428 085 063 68;

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:


0.000 000 000 742 147 676 645(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 645(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:


0.000 000 000 742 147 676 645(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 -0.000 000 000 742 147 676 645 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