-0.000 000 000 742 147 673 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 81(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
-0.000 000 000 742 147 673 81(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:

|-0.000 000 000 742 147 673 81| = 0.000 000 000 742 147 673 81


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: 0.000 000 000 742 147 673 81.

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) 0.000 000 000 742 147 673 81 × 2 = 0 + 0.000 000 001 484 295 347 62;
  • 2) 0.000 000 001 484 295 347 62 × 2 = 0 + 0.000 000 002 968 590 695 24;
  • 3) 0.000 000 002 968 590 695 24 × 2 = 0 + 0.000 000 005 937 181 390 48;
  • 4) 0.000 000 005 937 181 390 48 × 2 = 0 + 0.000 000 011 874 362 780 96;
  • 5) 0.000 000 011 874 362 780 96 × 2 = 0 + 0.000 000 023 748 725 561 92;
  • 6) 0.000 000 023 748 725 561 92 × 2 = 0 + 0.000 000 047 497 451 123 84;
  • 7) 0.000 000 047 497 451 123 84 × 2 = 0 + 0.000 000 094 994 902 247 68;
  • 8) 0.000 000 094 994 902 247 68 × 2 = 0 + 0.000 000 189 989 804 495 36;
  • 9) 0.000 000 189 989 804 495 36 × 2 = 0 + 0.000 000 379 979 608 990 72;
  • 10) 0.000 000 379 979 608 990 72 × 2 = 0 + 0.000 000 759 959 217 981 44;
  • 11) 0.000 000 759 959 217 981 44 × 2 = 0 + 0.000 001 519 918 435 962 88;
  • 12) 0.000 001 519 918 435 962 88 × 2 = 0 + 0.000 003 039 836 871 925 76;
  • 13) 0.000 003 039 836 871 925 76 × 2 = 0 + 0.000 006 079 673 743 851 52;
  • 14) 0.000 006 079 673 743 851 52 × 2 = 0 + 0.000 012 159 347 487 703 04;
  • 15) 0.000 012 159 347 487 703 04 × 2 = 0 + 0.000 024 318 694 975 406 08;
  • 16) 0.000 024 318 694 975 406 08 × 2 = 0 + 0.000 048 637 389 950 812 16;
  • 17) 0.000 048 637 389 950 812 16 × 2 = 0 + 0.000 097 274 779 901 624 32;
  • 18) 0.000 097 274 779 901 624 32 × 2 = 0 + 0.000 194 549 559 803 248 64;
  • 19) 0.000 194 549 559 803 248 64 × 2 = 0 + 0.000 389 099 119 606 497 28;
  • 20) 0.000 389 099 119 606 497 28 × 2 = 0 + 0.000 778 198 239 212 994 56;
  • 21) 0.000 778 198 239 212 994 56 × 2 = 0 + 0.001 556 396 478 425 989 12;
  • 22) 0.001 556 396 478 425 989 12 × 2 = 0 + 0.003 112 792 956 851 978 24;
  • 23) 0.003 112 792 956 851 978 24 × 2 = 0 + 0.006 225 585 913 703 956 48;
  • 24) 0.006 225 585 913 703 956 48 × 2 = 0 + 0.012 451 171 827 407 912 96;
  • 25) 0.012 451 171 827 407 912 96 × 2 = 0 + 0.024 902 343 654 815 825 92;
  • 26) 0.024 902 343 654 815 825 92 × 2 = 0 + 0.049 804 687 309 631 651 84;
  • 27) 0.049 804 687 309 631 651 84 × 2 = 0 + 0.099 609 374 619 263 303 68;
  • 28) 0.099 609 374 619 263 303 68 × 2 = 0 + 0.199 218 749 238 526 607 36;
  • 29) 0.199 218 749 238 526 607 36 × 2 = 0 + 0.398 437 498 477 053 214 72;
  • 30) 0.398 437 498 477 053 214 72 × 2 = 0 + 0.796 874 996 954 106 429 44;
  • 31) 0.796 874 996 954 106 429 44 × 2 = 1 + 0.593 749 993 908 212 858 88;
  • 32) 0.593 749 993 908 212 858 88 × 2 = 1 + 0.187 499 987 816 425 717 76;
  • 33) 0.187 499 987 816 425 717 76 × 2 = 0 + 0.374 999 975 632 851 435 52;
  • 34) 0.374 999 975 632 851 435 52 × 2 = 0 + 0.749 999 951 265 702 871 04;
  • 35) 0.749 999 951 265 702 871 04 × 2 = 1 + 0.499 999 902 531 405 742 08;
  • 36) 0.499 999 902 531 405 742 08 × 2 = 0 + 0.999 999 805 062 811 484 16;
  • 37) 0.999 999 805 062 811 484 16 × 2 = 1 + 0.999 999 610 125 622 968 32;
  • 38) 0.999 999 610 125 622 968 32 × 2 = 1 + 0.999 999 220 251 245 936 64;
  • 39) 0.999 999 220 251 245 936 64 × 2 = 1 + 0.999 998 440 502 491 873 28;
  • 40) 0.999 998 440 502 491 873 28 × 2 = 1 + 0.999 996 881 004 983 746 56;
  • 41) 0.999 996 881 004 983 746 56 × 2 = 1 + 0.999 993 762 009 967 493 12;
  • 42) 0.999 993 762 009 967 493 12 × 2 = 1 + 0.999 987 524 019 934 986 24;
  • 43) 0.999 987 524 019 934 986 24 × 2 = 1 + 0.999 975 048 039 869 972 48;
  • 44) 0.999 975 048 039 869 972 48 × 2 = 1 + 0.999 950 096 079 739 944 96;
  • 45) 0.999 950 096 079 739 944 96 × 2 = 1 + 0.999 900 192 159 479 889 92;
  • 46) 0.999 900 192 159 479 889 92 × 2 = 1 + 0.999 800 384 318 959 779 84;
  • 47) 0.999 800 384 318 959 779 84 × 2 = 1 + 0.999 600 768 637 919 559 68;
  • 48) 0.999 600 768 637 919 559 68 × 2 = 1 + 0.999 201 537 275 839 119 36;
  • 49) 0.999 201 537 275 839 119 36 × 2 = 1 + 0.998 403 074 551 678 238 72;
  • 50) 0.998 403 074 551 678 238 72 × 2 = 1 + 0.996 806 149 103 356 477 44;
  • 51) 0.996 806 149 103 356 477 44 × 2 = 1 + 0.993 612 298 206 712 954 88;
  • 52) 0.993 612 298 206 712 954 88 × 2 = 1 + 0.987 224 596 413 425 909 76;
  • 53) 0.987 224 596 413 425 909 76 × 2 = 1 + 0.974 449 192 826 851 819 52;
  • 54) 0.974 449 192 826 851 819 52 × 2 = 1 + 0.948 898 385 653 703 639 04;

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:


0.000 000 000 742 147 673 81(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 81(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:


0.000 000 000 742 147 673 81(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 -0.000 000 000 742 147 673 81 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