-0.000 000 000 742 147 676 481 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 481(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
-0.000 000 000 742 147 676 481(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:

|-0.000 000 000 742 147 676 481| = 0.000 000 000 742 147 676 481


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: 0.000 000 000 742 147 676 481.

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) 0.000 000 000 742 147 676 481 × 2 = 0 + 0.000 000 001 484 295 352 962;
  • 2) 0.000 000 001 484 295 352 962 × 2 = 0 + 0.000 000 002 968 590 705 924;
  • 3) 0.000 000 002 968 590 705 924 × 2 = 0 + 0.000 000 005 937 181 411 848;
  • 4) 0.000 000 005 937 181 411 848 × 2 = 0 + 0.000 000 011 874 362 823 696;
  • 5) 0.000 000 011 874 362 823 696 × 2 = 0 + 0.000 000 023 748 725 647 392;
  • 6) 0.000 000 023 748 725 647 392 × 2 = 0 + 0.000 000 047 497 451 294 784;
  • 7) 0.000 000 047 497 451 294 784 × 2 = 0 + 0.000 000 094 994 902 589 568;
  • 8) 0.000 000 094 994 902 589 568 × 2 = 0 + 0.000 000 189 989 805 179 136;
  • 9) 0.000 000 189 989 805 179 136 × 2 = 0 + 0.000 000 379 979 610 358 272;
  • 10) 0.000 000 379 979 610 358 272 × 2 = 0 + 0.000 000 759 959 220 716 544;
  • 11) 0.000 000 759 959 220 716 544 × 2 = 0 + 0.000 001 519 918 441 433 088;
  • 12) 0.000 001 519 918 441 433 088 × 2 = 0 + 0.000 003 039 836 882 866 176;
  • 13) 0.000 003 039 836 882 866 176 × 2 = 0 + 0.000 006 079 673 765 732 352;
  • 14) 0.000 006 079 673 765 732 352 × 2 = 0 + 0.000 012 159 347 531 464 704;
  • 15) 0.000 012 159 347 531 464 704 × 2 = 0 + 0.000 024 318 695 062 929 408;
  • 16) 0.000 024 318 695 062 929 408 × 2 = 0 + 0.000 048 637 390 125 858 816;
  • 17) 0.000 048 637 390 125 858 816 × 2 = 0 + 0.000 097 274 780 251 717 632;
  • 18) 0.000 097 274 780 251 717 632 × 2 = 0 + 0.000 194 549 560 503 435 264;
  • 19) 0.000 194 549 560 503 435 264 × 2 = 0 + 0.000 389 099 121 006 870 528;
  • 20) 0.000 389 099 121 006 870 528 × 2 = 0 + 0.000 778 198 242 013 741 056;
  • 21) 0.000 778 198 242 013 741 056 × 2 = 0 + 0.001 556 396 484 027 482 112;
  • 22) 0.001 556 396 484 027 482 112 × 2 = 0 + 0.003 112 792 968 054 964 224;
  • 23) 0.003 112 792 968 054 964 224 × 2 = 0 + 0.006 225 585 936 109 928 448;
  • 24) 0.006 225 585 936 109 928 448 × 2 = 0 + 0.012 451 171 872 219 856 896;
  • 25) 0.012 451 171 872 219 856 896 × 2 = 0 + 0.024 902 343 744 439 713 792;
  • 26) 0.024 902 343 744 439 713 792 × 2 = 0 + 0.049 804 687 488 879 427 584;
  • 27) 0.049 804 687 488 879 427 584 × 2 = 0 + 0.099 609 374 977 758 855 168;
  • 28) 0.099 609 374 977 758 855 168 × 2 = 0 + 0.199 218 749 955 517 710 336;
  • 29) 0.199 218 749 955 517 710 336 × 2 = 0 + 0.398 437 499 911 035 420 672;
  • 30) 0.398 437 499 911 035 420 672 × 2 = 0 + 0.796 874 999 822 070 841 344;
  • 31) 0.796 874 999 822 070 841 344 × 2 = 1 + 0.593 749 999 644 141 682 688;
  • 32) 0.593 749 999 644 141 682 688 × 2 = 1 + 0.187 499 999 288 283 365 376;
  • 33) 0.187 499 999 288 283 365 376 × 2 = 0 + 0.374 999 998 576 566 730 752;
  • 34) 0.374 999 998 576 566 730 752 × 2 = 0 + 0.749 999 997 153 133 461 504;
  • 35) 0.749 999 997 153 133 461 504 × 2 = 1 + 0.499 999 994 306 266 923 008;
  • 36) 0.499 999 994 306 266 923 008 × 2 = 0 + 0.999 999 988 612 533 846 016;
  • 37) 0.999 999 988 612 533 846 016 × 2 = 1 + 0.999 999 977 225 067 692 032;
  • 38) 0.999 999 977 225 067 692 032 × 2 = 1 + 0.999 999 954 450 135 384 064;
  • 39) 0.999 999 954 450 135 384 064 × 2 = 1 + 0.999 999 908 900 270 768 128;
  • 40) 0.999 999 908 900 270 768 128 × 2 = 1 + 0.999 999 817 800 541 536 256;
  • 41) 0.999 999 817 800 541 536 256 × 2 = 1 + 0.999 999 635 601 083 072 512;
  • 42) 0.999 999 635 601 083 072 512 × 2 = 1 + 0.999 999 271 202 166 145 024;
  • 43) 0.999 999 271 202 166 145 024 × 2 = 1 + 0.999 998 542 404 332 290 048;
  • 44) 0.999 998 542 404 332 290 048 × 2 = 1 + 0.999 997 084 808 664 580 096;
  • 45) 0.999 997 084 808 664 580 096 × 2 = 1 + 0.999 994 169 617 329 160 192;
  • 46) 0.999 994 169 617 329 160 192 × 2 = 1 + 0.999 988 339 234 658 320 384;
  • 47) 0.999 988 339 234 658 320 384 × 2 = 1 + 0.999 976 678 469 316 640 768;
  • 48) 0.999 976 678 469 316 640 768 × 2 = 1 + 0.999 953 356 938 633 281 536;
  • 49) 0.999 953 356 938 633 281 536 × 2 = 1 + 0.999 906 713 877 266 563 072;
  • 50) 0.999 906 713 877 266 563 072 × 2 = 1 + 0.999 813 427 754 533 126 144;
  • 51) 0.999 813 427 754 533 126 144 × 2 = 1 + 0.999 626 855 509 066 252 288;
  • 52) 0.999 626 855 509 066 252 288 × 2 = 1 + 0.999 253 711 018 132 504 576;
  • 53) 0.999 253 711 018 132 504 576 × 2 = 1 + 0.998 507 422 036 265 009 152;
  • 54) 0.998 507 422 036 265 009 152 × 2 = 1 + 0.997 014 844 072 530 018 304;

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:


0.000 000 000 742 147 676 481(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 481(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:


0.000 000 000 742 147 676 481(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 -0.000 000 000 742 147 676 481 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