-0.000 000 000 742 147 561 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 561(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
-0.000 000 000 742 147 561(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:

|-0.000 000 000 742 147 561| = 0.000 000 000 742 147 561


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: 0.000 000 000 742 147 561.

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) 0.000 000 000 742 147 561 × 2 = 0 + 0.000 000 001 484 295 122;
  • 2) 0.000 000 001 484 295 122 × 2 = 0 + 0.000 000 002 968 590 244;
  • 3) 0.000 000 002 968 590 244 × 2 = 0 + 0.000 000 005 937 180 488;
  • 4) 0.000 000 005 937 180 488 × 2 = 0 + 0.000 000 011 874 360 976;
  • 5) 0.000 000 011 874 360 976 × 2 = 0 + 0.000 000 023 748 721 952;
  • 6) 0.000 000 023 748 721 952 × 2 = 0 + 0.000 000 047 497 443 904;
  • 7) 0.000 000 047 497 443 904 × 2 = 0 + 0.000 000 094 994 887 808;
  • 8) 0.000 000 094 994 887 808 × 2 = 0 + 0.000 000 189 989 775 616;
  • 9) 0.000 000 189 989 775 616 × 2 = 0 + 0.000 000 379 979 551 232;
  • 10) 0.000 000 379 979 551 232 × 2 = 0 + 0.000 000 759 959 102 464;
  • 11) 0.000 000 759 959 102 464 × 2 = 0 + 0.000 001 519 918 204 928;
  • 12) 0.000 001 519 918 204 928 × 2 = 0 + 0.000 003 039 836 409 856;
  • 13) 0.000 003 039 836 409 856 × 2 = 0 + 0.000 006 079 672 819 712;
  • 14) 0.000 006 079 672 819 712 × 2 = 0 + 0.000 012 159 345 639 424;
  • 15) 0.000 012 159 345 639 424 × 2 = 0 + 0.000 024 318 691 278 848;
  • 16) 0.000 024 318 691 278 848 × 2 = 0 + 0.000 048 637 382 557 696;
  • 17) 0.000 048 637 382 557 696 × 2 = 0 + 0.000 097 274 765 115 392;
  • 18) 0.000 097 274 765 115 392 × 2 = 0 + 0.000 194 549 530 230 784;
  • 19) 0.000 194 549 530 230 784 × 2 = 0 + 0.000 389 099 060 461 568;
  • 20) 0.000 389 099 060 461 568 × 2 = 0 + 0.000 778 198 120 923 136;
  • 21) 0.000 778 198 120 923 136 × 2 = 0 + 0.001 556 396 241 846 272;
  • 22) 0.001 556 396 241 846 272 × 2 = 0 + 0.003 112 792 483 692 544;
  • 23) 0.003 112 792 483 692 544 × 2 = 0 + 0.006 225 584 967 385 088;
  • 24) 0.006 225 584 967 385 088 × 2 = 0 + 0.012 451 169 934 770 176;
  • 25) 0.012 451 169 934 770 176 × 2 = 0 + 0.024 902 339 869 540 352;
  • 26) 0.024 902 339 869 540 352 × 2 = 0 + 0.049 804 679 739 080 704;
  • 27) 0.049 804 679 739 080 704 × 2 = 0 + 0.099 609 359 478 161 408;
  • 28) 0.099 609 359 478 161 408 × 2 = 0 + 0.199 218 718 956 322 816;
  • 29) 0.199 218 718 956 322 816 × 2 = 0 + 0.398 437 437 912 645 632;
  • 30) 0.398 437 437 912 645 632 × 2 = 0 + 0.796 874 875 825 291 264;
  • 31) 0.796 874 875 825 291 264 × 2 = 1 + 0.593 749 751 650 582 528;
  • 32) 0.593 749 751 650 582 528 × 2 = 1 + 0.187 499 503 301 165 056;
  • 33) 0.187 499 503 301 165 056 × 2 = 0 + 0.374 999 006 602 330 112;
  • 34) 0.374 999 006 602 330 112 × 2 = 0 + 0.749 998 013 204 660 224;
  • 35) 0.749 998 013 204 660 224 × 2 = 1 + 0.499 996 026 409 320 448;
  • 36) 0.499 996 026 409 320 448 × 2 = 0 + 0.999 992 052 818 640 896;
  • 37) 0.999 992 052 818 640 896 × 2 = 1 + 0.999 984 105 637 281 792;
  • 38) 0.999 984 105 637 281 792 × 2 = 1 + 0.999 968 211 274 563 584;
  • 39) 0.999 968 211 274 563 584 × 2 = 1 + 0.999 936 422 549 127 168;
  • 40) 0.999 936 422 549 127 168 × 2 = 1 + 0.999 872 845 098 254 336;
  • 41) 0.999 872 845 098 254 336 × 2 = 1 + 0.999 745 690 196 508 672;
  • 42) 0.999 745 690 196 508 672 × 2 = 1 + 0.999 491 380 393 017 344;
  • 43) 0.999 491 380 393 017 344 × 2 = 1 + 0.998 982 760 786 034 688;
  • 44) 0.998 982 760 786 034 688 × 2 = 1 + 0.997 965 521 572 069 376;
  • 45) 0.997 965 521 572 069 376 × 2 = 1 + 0.995 931 043 144 138 752;
  • 46) 0.995 931 043 144 138 752 × 2 = 1 + 0.991 862 086 288 277 504;
  • 47) 0.991 862 086 288 277 504 × 2 = 1 + 0.983 724 172 576 555 008;
  • 48) 0.983 724 172 576 555 008 × 2 = 1 + 0.967 448 345 153 110 016;
  • 49) 0.967 448 345 153 110 016 × 2 = 1 + 0.934 896 690 306 220 032;
  • 50) 0.934 896 690 306 220 032 × 2 = 1 + 0.869 793 380 612 440 064;
  • 51) 0.869 793 380 612 440 064 × 2 = 1 + 0.739 586 761 224 880 128;
  • 52) 0.739 586 761 224 880 128 × 2 = 1 + 0.479 173 522 449 760 256;
  • 53) 0.479 173 522 449 760 256 × 2 = 0 + 0.958 347 044 899 520 512;
  • 54) 0.958 347 044 899 520 512 × 2 = 1 + 0.916 694 089 799 041 024;

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:


0.000 000 000 742 147 561(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 561(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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:


0.000 000 000 742 147 561(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 561 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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