-0.000 000 000 742 147 455 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 455(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
-0.000 000 000 742 147 455(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:

|-0.000 000 000 742 147 455| = 0.000 000 000 742 147 455


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: 0.000 000 000 742 147 455.

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) 0.000 000 000 742 147 455 × 2 = 0 + 0.000 000 001 484 294 91;
  • 2) 0.000 000 001 484 294 91 × 2 = 0 + 0.000 000 002 968 589 82;
  • 3) 0.000 000 002 968 589 82 × 2 = 0 + 0.000 000 005 937 179 64;
  • 4) 0.000 000 005 937 179 64 × 2 = 0 + 0.000 000 011 874 359 28;
  • 5) 0.000 000 011 874 359 28 × 2 = 0 + 0.000 000 023 748 718 56;
  • 6) 0.000 000 023 748 718 56 × 2 = 0 + 0.000 000 047 497 437 12;
  • 7) 0.000 000 047 497 437 12 × 2 = 0 + 0.000 000 094 994 874 24;
  • 8) 0.000 000 094 994 874 24 × 2 = 0 + 0.000 000 189 989 748 48;
  • 9) 0.000 000 189 989 748 48 × 2 = 0 + 0.000 000 379 979 496 96;
  • 10) 0.000 000 379 979 496 96 × 2 = 0 + 0.000 000 759 958 993 92;
  • 11) 0.000 000 759 958 993 92 × 2 = 0 + 0.000 001 519 917 987 84;
  • 12) 0.000 001 519 917 987 84 × 2 = 0 + 0.000 003 039 835 975 68;
  • 13) 0.000 003 039 835 975 68 × 2 = 0 + 0.000 006 079 671 951 36;
  • 14) 0.000 006 079 671 951 36 × 2 = 0 + 0.000 012 159 343 902 72;
  • 15) 0.000 012 159 343 902 72 × 2 = 0 + 0.000 024 318 687 805 44;
  • 16) 0.000 024 318 687 805 44 × 2 = 0 + 0.000 048 637 375 610 88;
  • 17) 0.000 048 637 375 610 88 × 2 = 0 + 0.000 097 274 751 221 76;
  • 18) 0.000 097 274 751 221 76 × 2 = 0 + 0.000 194 549 502 443 52;
  • 19) 0.000 194 549 502 443 52 × 2 = 0 + 0.000 389 099 004 887 04;
  • 20) 0.000 389 099 004 887 04 × 2 = 0 + 0.000 778 198 009 774 08;
  • 21) 0.000 778 198 009 774 08 × 2 = 0 + 0.001 556 396 019 548 16;
  • 22) 0.001 556 396 019 548 16 × 2 = 0 + 0.003 112 792 039 096 32;
  • 23) 0.003 112 792 039 096 32 × 2 = 0 + 0.006 225 584 078 192 64;
  • 24) 0.006 225 584 078 192 64 × 2 = 0 + 0.012 451 168 156 385 28;
  • 25) 0.012 451 168 156 385 28 × 2 = 0 + 0.024 902 336 312 770 56;
  • 26) 0.024 902 336 312 770 56 × 2 = 0 + 0.049 804 672 625 541 12;
  • 27) 0.049 804 672 625 541 12 × 2 = 0 + 0.099 609 345 251 082 24;
  • 28) 0.099 609 345 251 082 24 × 2 = 0 + 0.199 218 690 502 164 48;
  • 29) 0.199 218 690 502 164 48 × 2 = 0 + 0.398 437 381 004 328 96;
  • 30) 0.398 437 381 004 328 96 × 2 = 0 + 0.796 874 762 008 657 92;
  • 31) 0.796 874 762 008 657 92 × 2 = 1 + 0.593 749 524 017 315 84;
  • 32) 0.593 749 524 017 315 84 × 2 = 1 + 0.187 499 048 034 631 68;
  • 33) 0.187 499 048 034 631 68 × 2 = 0 + 0.374 998 096 069 263 36;
  • 34) 0.374 998 096 069 263 36 × 2 = 0 + 0.749 996 192 138 526 72;
  • 35) 0.749 996 192 138 526 72 × 2 = 1 + 0.499 992 384 277 053 44;
  • 36) 0.499 992 384 277 053 44 × 2 = 0 + 0.999 984 768 554 106 88;
  • 37) 0.999 984 768 554 106 88 × 2 = 1 + 0.999 969 537 108 213 76;
  • 38) 0.999 969 537 108 213 76 × 2 = 1 + 0.999 939 074 216 427 52;
  • 39) 0.999 939 074 216 427 52 × 2 = 1 + 0.999 878 148 432 855 04;
  • 40) 0.999 878 148 432 855 04 × 2 = 1 + 0.999 756 296 865 710 08;
  • 41) 0.999 756 296 865 710 08 × 2 = 1 + 0.999 512 593 731 420 16;
  • 42) 0.999 512 593 731 420 16 × 2 = 1 + 0.999 025 187 462 840 32;
  • 43) 0.999 025 187 462 840 32 × 2 = 1 + 0.998 050 374 925 680 64;
  • 44) 0.998 050 374 925 680 64 × 2 = 1 + 0.996 100 749 851 361 28;
  • 45) 0.996 100 749 851 361 28 × 2 = 1 + 0.992 201 499 702 722 56;
  • 46) 0.992 201 499 702 722 56 × 2 = 1 + 0.984 402 999 405 445 12;
  • 47) 0.984 402 999 405 445 12 × 2 = 1 + 0.968 805 998 810 890 24;
  • 48) 0.968 805 998 810 890 24 × 2 = 1 + 0.937 611 997 621 780 48;
  • 49) 0.937 611 997 621 780 48 × 2 = 1 + 0.875 223 995 243 560 96;
  • 50) 0.875 223 995 243 560 96 × 2 = 1 + 0.750 447 990 487 121 92;
  • 51) 0.750 447 990 487 121 92 × 2 = 1 + 0.500 895 980 974 243 84;
  • 52) 0.500 895 980 974 243 84 × 2 = 1 + 0.001 791 961 948 487 68;
  • 53) 0.001 791 961 948 487 68 × 2 = 0 + 0.003 583 923 896 975 36;
  • 54) 0.003 583 923 896 975 36 × 2 = 0 + 0.007 167 847 793 950 72;

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:


0.000 000 000 742 147 455(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 455(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 455(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 455 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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