-0.000 000 000 742 147 675 88 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 88(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 88(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 88| = 0.000 000 000 742 147 675 88


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 88.

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 88 × 2 = 0 + 0.000 000 001 484 295 351 76;
  • 2) 0.000 000 001 484 295 351 76 × 2 = 0 + 0.000 000 002 968 590 703 52;
  • 3) 0.000 000 002 968 590 703 52 × 2 = 0 + 0.000 000 005 937 181 407 04;
  • 4) 0.000 000 005 937 181 407 04 × 2 = 0 + 0.000 000 011 874 362 814 08;
  • 5) 0.000 000 011 874 362 814 08 × 2 = 0 + 0.000 000 023 748 725 628 16;
  • 6) 0.000 000 023 748 725 628 16 × 2 = 0 + 0.000 000 047 497 451 256 32;
  • 7) 0.000 000 047 497 451 256 32 × 2 = 0 + 0.000 000 094 994 902 512 64;
  • 8) 0.000 000 094 994 902 512 64 × 2 = 0 + 0.000 000 189 989 805 025 28;
  • 9) 0.000 000 189 989 805 025 28 × 2 = 0 + 0.000 000 379 979 610 050 56;
  • 10) 0.000 000 379 979 610 050 56 × 2 = 0 + 0.000 000 759 959 220 101 12;
  • 11) 0.000 000 759 959 220 101 12 × 2 = 0 + 0.000 001 519 918 440 202 24;
  • 12) 0.000 001 519 918 440 202 24 × 2 = 0 + 0.000 003 039 836 880 404 48;
  • 13) 0.000 003 039 836 880 404 48 × 2 = 0 + 0.000 006 079 673 760 808 96;
  • 14) 0.000 006 079 673 760 808 96 × 2 = 0 + 0.000 012 159 347 521 617 92;
  • 15) 0.000 012 159 347 521 617 92 × 2 = 0 + 0.000 024 318 695 043 235 84;
  • 16) 0.000 024 318 695 043 235 84 × 2 = 0 + 0.000 048 637 390 086 471 68;
  • 17) 0.000 048 637 390 086 471 68 × 2 = 0 + 0.000 097 274 780 172 943 36;
  • 18) 0.000 097 274 780 172 943 36 × 2 = 0 + 0.000 194 549 560 345 886 72;
  • 19) 0.000 194 549 560 345 886 72 × 2 = 0 + 0.000 389 099 120 691 773 44;
  • 20) 0.000 389 099 120 691 773 44 × 2 = 0 + 0.000 778 198 241 383 546 88;
  • 21) 0.000 778 198 241 383 546 88 × 2 = 0 + 0.001 556 396 482 767 093 76;
  • 22) 0.001 556 396 482 767 093 76 × 2 = 0 + 0.003 112 792 965 534 187 52;
  • 23) 0.003 112 792 965 534 187 52 × 2 = 0 + 0.006 225 585 931 068 375 04;
  • 24) 0.006 225 585 931 068 375 04 × 2 = 0 + 0.012 451 171 862 136 750 08;
  • 25) 0.012 451 171 862 136 750 08 × 2 = 0 + 0.024 902 343 724 273 500 16;
  • 26) 0.024 902 343 724 273 500 16 × 2 = 0 + 0.049 804 687 448 547 000 32;
  • 27) 0.049 804 687 448 547 000 32 × 2 = 0 + 0.099 609 374 897 094 000 64;
  • 28) 0.099 609 374 897 094 000 64 × 2 = 0 + 0.199 218 749 794 188 001 28;
  • 29) 0.199 218 749 794 188 001 28 × 2 = 0 + 0.398 437 499 588 376 002 56;
  • 30) 0.398 437 499 588 376 002 56 × 2 = 0 + 0.796 874 999 176 752 005 12;
  • 31) 0.796 874 999 176 752 005 12 × 2 = 1 + 0.593 749 998 353 504 010 24;
  • 32) 0.593 749 998 353 504 010 24 × 2 = 1 + 0.187 499 996 707 008 020 48;
  • 33) 0.187 499 996 707 008 020 48 × 2 = 0 + 0.374 999 993 414 016 040 96;
  • 34) 0.374 999 993 414 016 040 96 × 2 = 0 + 0.749 999 986 828 032 081 92;
  • 35) 0.749 999 986 828 032 081 92 × 2 = 1 + 0.499 999 973 656 064 163 84;
  • 36) 0.499 999 973 656 064 163 84 × 2 = 0 + 0.999 999 947 312 128 327 68;
  • 37) 0.999 999 947 312 128 327 68 × 2 = 1 + 0.999 999 894 624 256 655 36;
  • 38) 0.999 999 894 624 256 655 36 × 2 = 1 + 0.999 999 789 248 513 310 72;
  • 39) 0.999 999 789 248 513 310 72 × 2 = 1 + 0.999 999 578 497 026 621 44;
  • 40) 0.999 999 578 497 026 621 44 × 2 = 1 + 0.999 999 156 994 053 242 88;
  • 41) 0.999 999 156 994 053 242 88 × 2 = 1 + 0.999 998 313 988 106 485 76;
  • 42) 0.999 998 313 988 106 485 76 × 2 = 1 + 0.999 996 627 976 212 971 52;
  • 43) 0.999 996 627 976 212 971 52 × 2 = 1 + 0.999 993 255 952 425 943 04;
  • 44) 0.999 993 255 952 425 943 04 × 2 = 1 + 0.999 986 511 904 851 886 08;
  • 45) 0.999 986 511 904 851 886 08 × 2 = 1 + 0.999 973 023 809 703 772 16;
  • 46) 0.999 973 023 809 703 772 16 × 2 = 1 + 0.999 946 047 619 407 544 32;
  • 47) 0.999 946 047 619 407 544 32 × 2 = 1 + 0.999 892 095 238 815 088 64;
  • 48) 0.999 892 095 238 815 088 64 × 2 = 1 + 0.999 784 190 477 630 177 28;
  • 49) 0.999 784 190 477 630 177 28 × 2 = 1 + 0.999 568 380 955 260 354 56;
  • 50) 0.999 568 380 955 260 354 56 × 2 = 1 + 0.999 136 761 910 520 709 12;
  • 51) 0.999 136 761 910 520 709 12 × 2 = 1 + 0.998 273 523 821 041 418 24;
  • 52) 0.998 273 523 821 041 418 24 × 2 = 1 + 0.996 547 047 642 082 836 48;
  • 53) 0.996 547 047 642 082 836 48 × 2 = 1 + 0.993 094 095 284 165 672 96;
  • 54) 0.993 094 095 284 165 672 96 × 2 = 1 + 0.986 188 190 568 331 345 92;

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 88(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 88(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 88(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 88 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