-0.000 000 000 742 147 676 57 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 57(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 57(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 57| = 0.000 000 000 742 147 676 57


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 57.

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 57 × 2 = 0 + 0.000 000 001 484 295 353 14;
  • 2) 0.000 000 001 484 295 353 14 × 2 = 0 + 0.000 000 002 968 590 706 28;
  • 3) 0.000 000 002 968 590 706 28 × 2 = 0 + 0.000 000 005 937 181 412 56;
  • 4) 0.000 000 005 937 181 412 56 × 2 = 0 + 0.000 000 011 874 362 825 12;
  • 5) 0.000 000 011 874 362 825 12 × 2 = 0 + 0.000 000 023 748 725 650 24;
  • 6) 0.000 000 023 748 725 650 24 × 2 = 0 + 0.000 000 047 497 451 300 48;
  • 7) 0.000 000 047 497 451 300 48 × 2 = 0 + 0.000 000 094 994 902 600 96;
  • 8) 0.000 000 094 994 902 600 96 × 2 = 0 + 0.000 000 189 989 805 201 92;
  • 9) 0.000 000 189 989 805 201 92 × 2 = 0 + 0.000 000 379 979 610 403 84;
  • 10) 0.000 000 379 979 610 403 84 × 2 = 0 + 0.000 000 759 959 220 807 68;
  • 11) 0.000 000 759 959 220 807 68 × 2 = 0 + 0.000 001 519 918 441 615 36;
  • 12) 0.000 001 519 918 441 615 36 × 2 = 0 + 0.000 003 039 836 883 230 72;
  • 13) 0.000 003 039 836 883 230 72 × 2 = 0 + 0.000 006 079 673 766 461 44;
  • 14) 0.000 006 079 673 766 461 44 × 2 = 0 + 0.000 012 159 347 532 922 88;
  • 15) 0.000 012 159 347 532 922 88 × 2 = 0 + 0.000 024 318 695 065 845 76;
  • 16) 0.000 024 318 695 065 845 76 × 2 = 0 + 0.000 048 637 390 131 691 52;
  • 17) 0.000 048 637 390 131 691 52 × 2 = 0 + 0.000 097 274 780 263 383 04;
  • 18) 0.000 097 274 780 263 383 04 × 2 = 0 + 0.000 194 549 560 526 766 08;
  • 19) 0.000 194 549 560 526 766 08 × 2 = 0 + 0.000 389 099 121 053 532 16;
  • 20) 0.000 389 099 121 053 532 16 × 2 = 0 + 0.000 778 198 242 107 064 32;
  • 21) 0.000 778 198 242 107 064 32 × 2 = 0 + 0.001 556 396 484 214 128 64;
  • 22) 0.001 556 396 484 214 128 64 × 2 = 0 + 0.003 112 792 968 428 257 28;
  • 23) 0.003 112 792 968 428 257 28 × 2 = 0 + 0.006 225 585 936 856 514 56;
  • 24) 0.006 225 585 936 856 514 56 × 2 = 0 + 0.012 451 171 873 713 029 12;
  • 25) 0.012 451 171 873 713 029 12 × 2 = 0 + 0.024 902 343 747 426 058 24;
  • 26) 0.024 902 343 747 426 058 24 × 2 = 0 + 0.049 804 687 494 852 116 48;
  • 27) 0.049 804 687 494 852 116 48 × 2 = 0 + 0.099 609 374 989 704 232 96;
  • 28) 0.099 609 374 989 704 232 96 × 2 = 0 + 0.199 218 749 979 408 465 92;
  • 29) 0.199 218 749 979 408 465 92 × 2 = 0 + 0.398 437 499 958 816 931 84;
  • 30) 0.398 437 499 958 816 931 84 × 2 = 0 + 0.796 874 999 917 633 863 68;
  • 31) 0.796 874 999 917 633 863 68 × 2 = 1 + 0.593 749 999 835 267 727 36;
  • 32) 0.593 749 999 835 267 727 36 × 2 = 1 + 0.187 499 999 670 535 454 72;
  • 33) 0.187 499 999 670 535 454 72 × 2 = 0 + 0.374 999 999 341 070 909 44;
  • 34) 0.374 999 999 341 070 909 44 × 2 = 0 + 0.749 999 998 682 141 818 88;
  • 35) 0.749 999 998 682 141 818 88 × 2 = 1 + 0.499 999 997 364 283 637 76;
  • 36) 0.499 999 997 364 283 637 76 × 2 = 0 + 0.999 999 994 728 567 275 52;
  • 37) 0.999 999 994 728 567 275 52 × 2 = 1 + 0.999 999 989 457 134 551 04;
  • 38) 0.999 999 989 457 134 551 04 × 2 = 1 + 0.999 999 978 914 269 102 08;
  • 39) 0.999 999 978 914 269 102 08 × 2 = 1 + 0.999 999 957 828 538 204 16;
  • 40) 0.999 999 957 828 538 204 16 × 2 = 1 + 0.999 999 915 657 076 408 32;
  • 41) 0.999 999 915 657 076 408 32 × 2 = 1 + 0.999 999 831 314 152 816 64;
  • 42) 0.999 999 831 314 152 816 64 × 2 = 1 + 0.999 999 662 628 305 633 28;
  • 43) 0.999 999 662 628 305 633 28 × 2 = 1 + 0.999 999 325 256 611 266 56;
  • 44) 0.999 999 325 256 611 266 56 × 2 = 1 + 0.999 998 650 513 222 533 12;
  • 45) 0.999 998 650 513 222 533 12 × 2 = 1 + 0.999 997 301 026 445 066 24;
  • 46) 0.999 997 301 026 445 066 24 × 2 = 1 + 0.999 994 602 052 890 132 48;
  • 47) 0.999 994 602 052 890 132 48 × 2 = 1 + 0.999 989 204 105 780 264 96;
  • 48) 0.999 989 204 105 780 264 96 × 2 = 1 + 0.999 978 408 211 560 529 92;
  • 49) 0.999 978 408 211 560 529 92 × 2 = 1 + 0.999 956 816 423 121 059 84;
  • 50) 0.999 956 816 423 121 059 84 × 2 = 1 + 0.999 913 632 846 242 119 68;
  • 51) 0.999 913 632 846 242 119 68 × 2 = 1 + 0.999 827 265 692 484 239 36;
  • 52) 0.999 827 265 692 484 239 36 × 2 = 1 + 0.999 654 531 384 968 478 72;
  • 53) 0.999 654 531 384 968 478 72 × 2 = 1 + 0.999 309 062 769 936 957 44;
  • 54) 0.999 309 062 769 936 957 44 × 2 = 1 + 0.998 618 125 539 873 914 88;

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 57(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 57(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 57(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 57 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