-0.000 000 000 742 147 676 958 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 958(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
-0.000 000 000 742 147 676 958(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:

|-0.000 000 000 742 147 676 958| = 0.000 000 000 742 147 676 958


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: 0.000 000 000 742 147 676 958.

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) 0.000 000 000 742 147 676 958 × 2 = 0 + 0.000 000 001 484 295 353 916;
  • 2) 0.000 000 001 484 295 353 916 × 2 = 0 + 0.000 000 002 968 590 707 832;
  • 3) 0.000 000 002 968 590 707 832 × 2 = 0 + 0.000 000 005 937 181 415 664;
  • 4) 0.000 000 005 937 181 415 664 × 2 = 0 + 0.000 000 011 874 362 831 328;
  • 5) 0.000 000 011 874 362 831 328 × 2 = 0 + 0.000 000 023 748 725 662 656;
  • 6) 0.000 000 023 748 725 662 656 × 2 = 0 + 0.000 000 047 497 451 325 312;
  • 7) 0.000 000 047 497 451 325 312 × 2 = 0 + 0.000 000 094 994 902 650 624;
  • 8) 0.000 000 094 994 902 650 624 × 2 = 0 + 0.000 000 189 989 805 301 248;
  • 9) 0.000 000 189 989 805 301 248 × 2 = 0 + 0.000 000 379 979 610 602 496;
  • 10) 0.000 000 379 979 610 602 496 × 2 = 0 + 0.000 000 759 959 221 204 992;
  • 11) 0.000 000 759 959 221 204 992 × 2 = 0 + 0.000 001 519 918 442 409 984;
  • 12) 0.000 001 519 918 442 409 984 × 2 = 0 + 0.000 003 039 836 884 819 968;
  • 13) 0.000 003 039 836 884 819 968 × 2 = 0 + 0.000 006 079 673 769 639 936;
  • 14) 0.000 006 079 673 769 639 936 × 2 = 0 + 0.000 012 159 347 539 279 872;
  • 15) 0.000 012 159 347 539 279 872 × 2 = 0 + 0.000 024 318 695 078 559 744;
  • 16) 0.000 024 318 695 078 559 744 × 2 = 0 + 0.000 048 637 390 157 119 488;
  • 17) 0.000 048 637 390 157 119 488 × 2 = 0 + 0.000 097 274 780 314 238 976;
  • 18) 0.000 097 274 780 314 238 976 × 2 = 0 + 0.000 194 549 560 628 477 952;
  • 19) 0.000 194 549 560 628 477 952 × 2 = 0 + 0.000 389 099 121 256 955 904;
  • 20) 0.000 389 099 121 256 955 904 × 2 = 0 + 0.000 778 198 242 513 911 808;
  • 21) 0.000 778 198 242 513 911 808 × 2 = 0 + 0.001 556 396 485 027 823 616;
  • 22) 0.001 556 396 485 027 823 616 × 2 = 0 + 0.003 112 792 970 055 647 232;
  • 23) 0.003 112 792 970 055 647 232 × 2 = 0 + 0.006 225 585 940 111 294 464;
  • 24) 0.006 225 585 940 111 294 464 × 2 = 0 + 0.012 451 171 880 222 588 928;
  • 25) 0.012 451 171 880 222 588 928 × 2 = 0 + 0.024 902 343 760 445 177 856;
  • 26) 0.024 902 343 760 445 177 856 × 2 = 0 + 0.049 804 687 520 890 355 712;
  • 27) 0.049 804 687 520 890 355 712 × 2 = 0 + 0.099 609 375 041 780 711 424;
  • 28) 0.099 609 375 041 780 711 424 × 2 = 0 + 0.199 218 750 083 561 422 848;
  • 29) 0.199 218 750 083 561 422 848 × 2 = 0 + 0.398 437 500 167 122 845 696;
  • 30) 0.398 437 500 167 122 845 696 × 2 = 0 + 0.796 875 000 334 245 691 392;
  • 31) 0.796 875 000 334 245 691 392 × 2 = 1 + 0.593 750 000 668 491 382 784;
  • 32) 0.593 750 000 668 491 382 784 × 2 = 1 + 0.187 500 001 336 982 765 568;
  • 33) 0.187 500 001 336 982 765 568 × 2 = 0 + 0.375 000 002 673 965 531 136;
  • 34) 0.375 000 002 673 965 531 136 × 2 = 0 + 0.750 000 005 347 931 062 272;
  • 35) 0.750 000 005 347 931 062 272 × 2 = 1 + 0.500 000 010 695 862 124 544;
  • 36) 0.500 000 010 695 862 124 544 × 2 = 1 + 0.000 000 021 391 724 249 088;
  • 37) 0.000 000 021 391 724 249 088 × 2 = 0 + 0.000 000 042 783 448 498 176;
  • 38) 0.000 000 042 783 448 498 176 × 2 = 0 + 0.000 000 085 566 896 996 352;
  • 39) 0.000 000 085 566 896 996 352 × 2 = 0 + 0.000 000 171 133 793 992 704;
  • 40) 0.000 000 171 133 793 992 704 × 2 = 0 + 0.000 000 342 267 587 985 408;
  • 41) 0.000 000 342 267 587 985 408 × 2 = 0 + 0.000 000 684 535 175 970 816;
  • 42) 0.000 000 684 535 175 970 816 × 2 = 0 + 0.000 001 369 070 351 941 632;
  • 43) 0.000 001 369 070 351 941 632 × 2 = 0 + 0.000 002 738 140 703 883 264;
  • 44) 0.000 002 738 140 703 883 264 × 2 = 0 + 0.000 005 476 281 407 766 528;
  • 45) 0.000 005 476 281 407 766 528 × 2 = 0 + 0.000 010 952 562 815 533 056;
  • 46) 0.000 010 952 562 815 533 056 × 2 = 0 + 0.000 021 905 125 631 066 112;
  • 47) 0.000 021 905 125 631 066 112 × 2 = 0 + 0.000 043 810 251 262 132 224;
  • 48) 0.000 043 810 251 262 132 224 × 2 = 0 + 0.000 087 620 502 524 264 448;
  • 49) 0.000 087 620 502 524 264 448 × 2 = 0 + 0.000 175 241 005 048 528 896;
  • 50) 0.000 175 241 005 048 528 896 × 2 = 0 + 0.000 350 482 010 097 057 792;
  • 51) 0.000 350 482 010 097 057 792 × 2 = 0 + 0.000 700 964 020 194 115 584;
  • 52) 0.000 700 964 020 194 115 584 × 2 = 0 + 0.001 401 928 040 388 231 168;
  • 53) 0.001 401 928 040 388 231 168 × 2 = 0 + 0.002 803 856 080 776 462 336;
  • 54) 0.002 803 856 080 776 462 336 × 2 = 0 + 0.005 607 712 161 552 924 672;

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:


0.000 000 000 742 147 676 958(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 958(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:


0.000 000 000 742 147 676 958(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 -0.000 000 000 742 147 676 958 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