-0.000 000 000 742 147 676 01 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 01(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 01(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 01| = 0.000 000 000 742 147 676 01


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 01.

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 01 × 2 = 0 + 0.000 000 001 484 295 352 02;
  • 2) 0.000 000 001 484 295 352 02 × 2 = 0 + 0.000 000 002 968 590 704 04;
  • 3) 0.000 000 002 968 590 704 04 × 2 = 0 + 0.000 000 005 937 181 408 08;
  • 4) 0.000 000 005 937 181 408 08 × 2 = 0 + 0.000 000 011 874 362 816 16;
  • 5) 0.000 000 011 874 362 816 16 × 2 = 0 + 0.000 000 023 748 725 632 32;
  • 6) 0.000 000 023 748 725 632 32 × 2 = 0 + 0.000 000 047 497 451 264 64;
  • 7) 0.000 000 047 497 451 264 64 × 2 = 0 + 0.000 000 094 994 902 529 28;
  • 8) 0.000 000 094 994 902 529 28 × 2 = 0 + 0.000 000 189 989 805 058 56;
  • 9) 0.000 000 189 989 805 058 56 × 2 = 0 + 0.000 000 379 979 610 117 12;
  • 10) 0.000 000 379 979 610 117 12 × 2 = 0 + 0.000 000 759 959 220 234 24;
  • 11) 0.000 000 759 959 220 234 24 × 2 = 0 + 0.000 001 519 918 440 468 48;
  • 12) 0.000 001 519 918 440 468 48 × 2 = 0 + 0.000 003 039 836 880 936 96;
  • 13) 0.000 003 039 836 880 936 96 × 2 = 0 + 0.000 006 079 673 761 873 92;
  • 14) 0.000 006 079 673 761 873 92 × 2 = 0 + 0.000 012 159 347 523 747 84;
  • 15) 0.000 012 159 347 523 747 84 × 2 = 0 + 0.000 024 318 695 047 495 68;
  • 16) 0.000 024 318 695 047 495 68 × 2 = 0 + 0.000 048 637 390 094 991 36;
  • 17) 0.000 048 637 390 094 991 36 × 2 = 0 + 0.000 097 274 780 189 982 72;
  • 18) 0.000 097 274 780 189 982 72 × 2 = 0 + 0.000 194 549 560 379 965 44;
  • 19) 0.000 194 549 560 379 965 44 × 2 = 0 + 0.000 389 099 120 759 930 88;
  • 20) 0.000 389 099 120 759 930 88 × 2 = 0 + 0.000 778 198 241 519 861 76;
  • 21) 0.000 778 198 241 519 861 76 × 2 = 0 + 0.001 556 396 483 039 723 52;
  • 22) 0.001 556 396 483 039 723 52 × 2 = 0 + 0.003 112 792 966 079 447 04;
  • 23) 0.003 112 792 966 079 447 04 × 2 = 0 + 0.006 225 585 932 158 894 08;
  • 24) 0.006 225 585 932 158 894 08 × 2 = 0 + 0.012 451 171 864 317 788 16;
  • 25) 0.012 451 171 864 317 788 16 × 2 = 0 + 0.024 902 343 728 635 576 32;
  • 26) 0.024 902 343 728 635 576 32 × 2 = 0 + 0.049 804 687 457 271 152 64;
  • 27) 0.049 804 687 457 271 152 64 × 2 = 0 + 0.099 609 374 914 542 305 28;
  • 28) 0.099 609 374 914 542 305 28 × 2 = 0 + 0.199 218 749 829 084 610 56;
  • 29) 0.199 218 749 829 084 610 56 × 2 = 0 + 0.398 437 499 658 169 221 12;
  • 30) 0.398 437 499 658 169 221 12 × 2 = 0 + 0.796 874 999 316 338 442 24;
  • 31) 0.796 874 999 316 338 442 24 × 2 = 1 + 0.593 749 998 632 676 884 48;
  • 32) 0.593 749 998 632 676 884 48 × 2 = 1 + 0.187 499 997 265 353 768 96;
  • 33) 0.187 499 997 265 353 768 96 × 2 = 0 + 0.374 999 994 530 707 537 92;
  • 34) 0.374 999 994 530 707 537 92 × 2 = 0 + 0.749 999 989 061 415 075 84;
  • 35) 0.749 999 989 061 415 075 84 × 2 = 1 + 0.499 999 978 122 830 151 68;
  • 36) 0.499 999 978 122 830 151 68 × 2 = 0 + 0.999 999 956 245 660 303 36;
  • 37) 0.999 999 956 245 660 303 36 × 2 = 1 + 0.999 999 912 491 320 606 72;
  • 38) 0.999 999 912 491 320 606 72 × 2 = 1 + 0.999 999 824 982 641 213 44;
  • 39) 0.999 999 824 982 641 213 44 × 2 = 1 + 0.999 999 649 965 282 426 88;
  • 40) 0.999 999 649 965 282 426 88 × 2 = 1 + 0.999 999 299 930 564 853 76;
  • 41) 0.999 999 299 930 564 853 76 × 2 = 1 + 0.999 998 599 861 129 707 52;
  • 42) 0.999 998 599 861 129 707 52 × 2 = 1 + 0.999 997 199 722 259 415 04;
  • 43) 0.999 997 199 722 259 415 04 × 2 = 1 + 0.999 994 399 444 518 830 08;
  • 44) 0.999 994 399 444 518 830 08 × 2 = 1 + 0.999 988 798 889 037 660 16;
  • 45) 0.999 988 798 889 037 660 16 × 2 = 1 + 0.999 977 597 778 075 320 32;
  • 46) 0.999 977 597 778 075 320 32 × 2 = 1 + 0.999 955 195 556 150 640 64;
  • 47) 0.999 955 195 556 150 640 64 × 2 = 1 + 0.999 910 391 112 301 281 28;
  • 48) 0.999 910 391 112 301 281 28 × 2 = 1 + 0.999 820 782 224 602 562 56;
  • 49) 0.999 820 782 224 602 562 56 × 2 = 1 + 0.999 641 564 449 205 125 12;
  • 50) 0.999 641 564 449 205 125 12 × 2 = 1 + 0.999 283 128 898 410 250 24;
  • 51) 0.999 283 128 898 410 250 24 × 2 = 1 + 0.998 566 257 796 820 500 48;
  • 52) 0.998 566 257 796 820 500 48 × 2 = 1 + 0.997 132 515 593 641 000 96;
  • 53) 0.997 132 515 593 641 000 96 × 2 = 1 + 0.994 265 031 187 282 001 92;
  • 54) 0.994 265 031 187 282 001 92 × 2 = 1 + 0.988 530 062 374 564 003 84;

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 01(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 01(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 01(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 01 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