-0.000 000 000 742 147 675 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 93(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
-0.000 000 000 742 147 675 93(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:

|-0.000 000 000 742 147 675 93| = 0.000 000 000 742 147 675 93


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: 0.000 000 000 742 147 675 93.

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) 0.000 000 000 742 147 675 93 × 2 = 0 + 0.000 000 001 484 295 351 86;
  • 2) 0.000 000 001 484 295 351 86 × 2 = 0 + 0.000 000 002 968 590 703 72;
  • 3) 0.000 000 002 968 590 703 72 × 2 = 0 + 0.000 000 005 937 181 407 44;
  • 4) 0.000 000 005 937 181 407 44 × 2 = 0 + 0.000 000 011 874 362 814 88;
  • 5) 0.000 000 011 874 362 814 88 × 2 = 0 + 0.000 000 023 748 725 629 76;
  • 6) 0.000 000 023 748 725 629 76 × 2 = 0 + 0.000 000 047 497 451 259 52;
  • 7) 0.000 000 047 497 451 259 52 × 2 = 0 + 0.000 000 094 994 902 519 04;
  • 8) 0.000 000 094 994 902 519 04 × 2 = 0 + 0.000 000 189 989 805 038 08;
  • 9) 0.000 000 189 989 805 038 08 × 2 = 0 + 0.000 000 379 979 610 076 16;
  • 10) 0.000 000 379 979 610 076 16 × 2 = 0 + 0.000 000 759 959 220 152 32;
  • 11) 0.000 000 759 959 220 152 32 × 2 = 0 + 0.000 001 519 918 440 304 64;
  • 12) 0.000 001 519 918 440 304 64 × 2 = 0 + 0.000 003 039 836 880 609 28;
  • 13) 0.000 003 039 836 880 609 28 × 2 = 0 + 0.000 006 079 673 761 218 56;
  • 14) 0.000 006 079 673 761 218 56 × 2 = 0 + 0.000 012 159 347 522 437 12;
  • 15) 0.000 012 159 347 522 437 12 × 2 = 0 + 0.000 024 318 695 044 874 24;
  • 16) 0.000 024 318 695 044 874 24 × 2 = 0 + 0.000 048 637 390 089 748 48;
  • 17) 0.000 048 637 390 089 748 48 × 2 = 0 + 0.000 097 274 780 179 496 96;
  • 18) 0.000 097 274 780 179 496 96 × 2 = 0 + 0.000 194 549 560 358 993 92;
  • 19) 0.000 194 549 560 358 993 92 × 2 = 0 + 0.000 389 099 120 717 987 84;
  • 20) 0.000 389 099 120 717 987 84 × 2 = 0 + 0.000 778 198 241 435 975 68;
  • 21) 0.000 778 198 241 435 975 68 × 2 = 0 + 0.001 556 396 482 871 951 36;
  • 22) 0.001 556 396 482 871 951 36 × 2 = 0 + 0.003 112 792 965 743 902 72;
  • 23) 0.003 112 792 965 743 902 72 × 2 = 0 + 0.006 225 585 931 487 805 44;
  • 24) 0.006 225 585 931 487 805 44 × 2 = 0 + 0.012 451 171 862 975 610 88;
  • 25) 0.012 451 171 862 975 610 88 × 2 = 0 + 0.024 902 343 725 951 221 76;
  • 26) 0.024 902 343 725 951 221 76 × 2 = 0 + 0.049 804 687 451 902 443 52;
  • 27) 0.049 804 687 451 902 443 52 × 2 = 0 + 0.099 609 374 903 804 887 04;
  • 28) 0.099 609 374 903 804 887 04 × 2 = 0 + 0.199 218 749 807 609 774 08;
  • 29) 0.199 218 749 807 609 774 08 × 2 = 0 + 0.398 437 499 615 219 548 16;
  • 30) 0.398 437 499 615 219 548 16 × 2 = 0 + 0.796 874 999 230 439 096 32;
  • 31) 0.796 874 999 230 439 096 32 × 2 = 1 + 0.593 749 998 460 878 192 64;
  • 32) 0.593 749 998 460 878 192 64 × 2 = 1 + 0.187 499 996 921 756 385 28;
  • 33) 0.187 499 996 921 756 385 28 × 2 = 0 + 0.374 999 993 843 512 770 56;
  • 34) 0.374 999 993 843 512 770 56 × 2 = 0 + 0.749 999 987 687 025 541 12;
  • 35) 0.749 999 987 687 025 541 12 × 2 = 1 + 0.499 999 975 374 051 082 24;
  • 36) 0.499 999 975 374 051 082 24 × 2 = 0 + 0.999 999 950 748 102 164 48;
  • 37) 0.999 999 950 748 102 164 48 × 2 = 1 + 0.999 999 901 496 204 328 96;
  • 38) 0.999 999 901 496 204 328 96 × 2 = 1 + 0.999 999 802 992 408 657 92;
  • 39) 0.999 999 802 992 408 657 92 × 2 = 1 + 0.999 999 605 984 817 315 84;
  • 40) 0.999 999 605 984 817 315 84 × 2 = 1 + 0.999 999 211 969 634 631 68;
  • 41) 0.999 999 211 969 634 631 68 × 2 = 1 + 0.999 998 423 939 269 263 36;
  • 42) 0.999 998 423 939 269 263 36 × 2 = 1 + 0.999 996 847 878 538 526 72;
  • 43) 0.999 996 847 878 538 526 72 × 2 = 1 + 0.999 993 695 757 077 053 44;
  • 44) 0.999 993 695 757 077 053 44 × 2 = 1 + 0.999 987 391 514 154 106 88;
  • 45) 0.999 987 391 514 154 106 88 × 2 = 1 + 0.999 974 783 028 308 213 76;
  • 46) 0.999 974 783 028 308 213 76 × 2 = 1 + 0.999 949 566 056 616 427 52;
  • 47) 0.999 949 566 056 616 427 52 × 2 = 1 + 0.999 899 132 113 232 855 04;
  • 48) 0.999 899 132 113 232 855 04 × 2 = 1 + 0.999 798 264 226 465 710 08;
  • 49) 0.999 798 264 226 465 710 08 × 2 = 1 + 0.999 596 528 452 931 420 16;
  • 50) 0.999 596 528 452 931 420 16 × 2 = 1 + 0.999 193 056 905 862 840 32;
  • 51) 0.999 193 056 905 862 840 32 × 2 = 1 + 0.998 386 113 811 725 680 64;
  • 52) 0.998 386 113 811 725 680 64 × 2 = 1 + 0.996 772 227 623 451 361 28;
  • 53) 0.996 772 227 623 451 361 28 × 2 = 1 + 0.993 544 455 246 902 722 56;
  • 54) 0.993 544 455 246 902 722 56 × 2 = 1 + 0.987 088 910 493 805 445 12;

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:


0.000 000 000 742 147 675 93(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 93(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:


0.000 000 000 742 147 675 93(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 -0.000 000 000 742 147 675 93 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