-0.000 000 000 742 147 676 646 717 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 717 5(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
-0.000 000 000 742 147 676 646 717 5(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:

|-0.000 000 000 742 147 676 646 717 5| = 0.000 000 000 742 147 676 646 717 5


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: 0.000 000 000 742 147 676 646 717 5.

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) 0.000 000 000 742 147 676 646 717 5 × 2 = 0 + 0.000 000 001 484 295 353 293 435;
  • 2) 0.000 000 001 484 295 353 293 435 × 2 = 0 + 0.000 000 002 968 590 706 586 87;
  • 3) 0.000 000 002 968 590 706 586 87 × 2 = 0 + 0.000 000 005 937 181 413 173 74;
  • 4) 0.000 000 005 937 181 413 173 74 × 2 = 0 + 0.000 000 011 874 362 826 347 48;
  • 5) 0.000 000 011 874 362 826 347 48 × 2 = 0 + 0.000 000 023 748 725 652 694 96;
  • 6) 0.000 000 023 748 725 652 694 96 × 2 = 0 + 0.000 000 047 497 451 305 389 92;
  • 7) 0.000 000 047 497 451 305 389 92 × 2 = 0 + 0.000 000 094 994 902 610 779 84;
  • 8) 0.000 000 094 994 902 610 779 84 × 2 = 0 + 0.000 000 189 989 805 221 559 68;
  • 9) 0.000 000 189 989 805 221 559 68 × 2 = 0 + 0.000 000 379 979 610 443 119 36;
  • 10) 0.000 000 379 979 610 443 119 36 × 2 = 0 + 0.000 000 759 959 220 886 238 72;
  • 11) 0.000 000 759 959 220 886 238 72 × 2 = 0 + 0.000 001 519 918 441 772 477 44;
  • 12) 0.000 001 519 918 441 772 477 44 × 2 = 0 + 0.000 003 039 836 883 544 954 88;
  • 13) 0.000 003 039 836 883 544 954 88 × 2 = 0 + 0.000 006 079 673 767 089 909 76;
  • 14) 0.000 006 079 673 767 089 909 76 × 2 = 0 + 0.000 012 159 347 534 179 819 52;
  • 15) 0.000 012 159 347 534 179 819 52 × 2 = 0 + 0.000 024 318 695 068 359 639 04;
  • 16) 0.000 024 318 695 068 359 639 04 × 2 = 0 + 0.000 048 637 390 136 719 278 08;
  • 17) 0.000 048 637 390 136 719 278 08 × 2 = 0 + 0.000 097 274 780 273 438 556 16;
  • 18) 0.000 097 274 780 273 438 556 16 × 2 = 0 + 0.000 194 549 560 546 877 112 32;
  • 19) 0.000 194 549 560 546 877 112 32 × 2 = 0 + 0.000 389 099 121 093 754 224 64;
  • 20) 0.000 389 099 121 093 754 224 64 × 2 = 0 + 0.000 778 198 242 187 508 449 28;
  • 21) 0.000 778 198 242 187 508 449 28 × 2 = 0 + 0.001 556 396 484 375 016 898 56;
  • 22) 0.001 556 396 484 375 016 898 56 × 2 = 0 + 0.003 112 792 968 750 033 797 12;
  • 23) 0.003 112 792 968 750 033 797 12 × 2 = 0 + 0.006 225 585 937 500 067 594 24;
  • 24) 0.006 225 585 937 500 067 594 24 × 2 = 0 + 0.012 451 171 875 000 135 188 48;
  • 25) 0.012 451 171 875 000 135 188 48 × 2 = 0 + 0.024 902 343 750 000 270 376 96;
  • 26) 0.024 902 343 750 000 270 376 96 × 2 = 0 + 0.049 804 687 500 000 540 753 92;
  • 27) 0.049 804 687 500 000 540 753 92 × 2 = 0 + 0.099 609 375 000 001 081 507 84;
  • 28) 0.099 609 375 000 001 081 507 84 × 2 = 0 + 0.199 218 750 000 002 163 015 68;
  • 29) 0.199 218 750 000 002 163 015 68 × 2 = 0 + 0.398 437 500 000 004 326 031 36;
  • 30) 0.398 437 500 000 004 326 031 36 × 2 = 0 + 0.796 875 000 000 008 652 062 72;
  • 31) 0.796 875 000 000 008 652 062 72 × 2 = 1 + 0.593 750 000 000 017 304 125 44;
  • 32) 0.593 750 000 000 017 304 125 44 × 2 = 1 + 0.187 500 000 000 034 608 250 88;
  • 33) 0.187 500 000 000 034 608 250 88 × 2 = 0 + 0.375 000 000 000 069 216 501 76;
  • 34) 0.375 000 000 000 069 216 501 76 × 2 = 0 + 0.750 000 000 000 138 433 003 52;
  • 35) 0.750 000 000 000 138 433 003 52 × 2 = 1 + 0.500 000 000 000 276 866 007 04;
  • 36) 0.500 000 000 000 276 866 007 04 × 2 = 1 + 0.000 000 000 000 553 732 014 08;
  • 37) 0.000 000 000 000 553 732 014 08 × 2 = 0 + 0.000 000 000 001 107 464 028 16;
  • 38) 0.000 000 000 001 107 464 028 16 × 2 = 0 + 0.000 000 000 002 214 928 056 32;
  • 39) 0.000 000 000 002 214 928 056 32 × 2 = 0 + 0.000 000 000 004 429 856 112 64;
  • 40) 0.000 000 000 004 429 856 112 64 × 2 = 0 + 0.000 000 000 008 859 712 225 28;
  • 41) 0.000 000 000 008 859 712 225 28 × 2 = 0 + 0.000 000 000 017 719 424 450 56;
  • 42) 0.000 000 000 017 719 424 450 56 × 2 = 0 + 0.000 000 000 035 438 848 901 12;
  • 43) 0.000 000 000 035 438 848 901 12 × 2 = 0 + 0.000 000 000 070 877 697 802 24;
  • 44) 0.000 000 000 070 877 697 802 24 × 2 = 0 + 0.000 000 000 141 755 395 604 48;
  • 45) 0.000 000 000 141 755 395 604 48 × 2 = 0 + 0.000 000 000 283 510 791 208 96;
  • 46) 0.000 000 000 283 510 791 208 96 × 2 = 0 + 0.000 000 000 567 021 582 417 92;
  • 47) 0.000 000 000 567 021 582 417 92 × 2 = 0 + 0.000 000 001 134 043 164 835 84;
  • 48) 0.000 000 001 134 043 164 835 84 × 2 = 0 + 0.000 000 002 268 086 329 671 68;
  • 49) 0.000 000 002 268 086 329 671 68 × 2 = 0 + 0.000 000 004 536 172 659 343 36;
  • 50) 0.000 000 004 536 172 659 343 36 × 2 = 0 + 0.000 000 009 072 345 318 686 72;
  • 51) 0.000 000 009 072 345 318 686 72 × 2 = 0 + 0.000 000 018 144 690 637 373 44;
  • 52) 0.000 000 018 144 690 637 373 44 × 2 = 0 + 0.000 000 036 289 381 274 746 88;
  • 53) 0.000 000 036 289 381 274 746 88 × 2 = 0 + 0.000 000 072 578 762 549 493 76;
  • 54) 0.000 000 072 578 762 549 493 76 × 2 = 0 + 0.000 000 145 157 525 098 987 52;

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:


0.000 000 000 742 147 676 646 717 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 717 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 676 646 717 5(10) =

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

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

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 717 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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