-0.000 000 000 742 147 811 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 811(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
-0.000 000 000 742 147 811(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:

|-0.000 000 000 742 147 811| = 0.000 000 000 742 147 811


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: 0.000 000 000 742 147 811.

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) 0.000 000 000 742 147 811 × 2 = 0 + 0.000 000 001 484 295 622;
  • 2) 0.000 000 001 484 295 622 × 2 = 0 + 0.000 000 002 968 591 244;
  • 3) 0.000 000 002 968 591 244 × 2 = 0 + 0.000 000 005 937 182 488;
  • 4) 0.000 000 005 937 182 488 × 2 = 0 + 0.000 000 011 874 364 976;
  • 5) 0.000 000 011 874 364 976 × 2 = 0 + 0.000 000 023 748 729 952;
  • 6) 0.000 000 023 748 729 952 × 2 = 0 + 0.000 000 047 497 459 904;
  • 7) 0.000 000 047 497 459 904 × 2 = 0 + 0.000 000 094 994 919 808;
  • 8) 0.000 000 094 994 919 808 × 2 = 0 + 0.000 000 189 989 839 616;
  • 9) 0.000 000 189 989 839 616 × 2 = 0 + 0.000 000 379 979 679 232;
  • 10) 0.000 000 379 979 679 232 × 2 = 0 + 0.000 000 759 959 358 464;
  • 11) 0.000 000 759 959 358 464 × 2 = 0 + 0.000 001 519 918 716 928;
  • 12) 0.000 001 519 918 716 928 × 2 = 0 + 0.000 003 039 837 433 856;
  • 13) 0.000 003 039 837 433 856 × 2 = 0 + 0.000 006 079 674 867 712;
  • 14) 0.000 006 079 674 867 712 × 2 = 0 + 0.000 012 159 349 735 424;
  • 15) 0.000 012 159 349 735 424 × 2 = 0 + 0.000 024 318 699 470 848;
  • 16) 0.000 024 318 699 470 848 × 2 = 0 + 0.000 048 637 398 941 696;
  • 17) 0.000 048 637 398 941 696 × 2 = 0 + 0.000 097 274 797 883 392;
  • 18) 0.000 097 274 797 883 392 × 2 = 0 + 0.000 194 549 595 766 784;
  • 19) 0.000 194 549 595 766 784 × 2 = 0 + 0.000 389 099 191 533 568;
  • 20) 0.000 389 099 191 533 568 × 2 = 0 + 0.000 778 198 383 067 136;
  • 21) 0.000 778 198 383 067 136 × 2 = 0 + 0.001 556 396 766 134 272;
  • 22) 0.001 556 396 766 134 272 × 2 = 0 + 0.003 112 793 532 268 544;
  • 23) 0.003 112 793 532 268 544 × 2 = 0 + 0.006 225 587 064 537 088;
  • 24) 0.006 225 587 064 537 088 × 2 = 0 + 0.012 451 174 129 074 176;
  • 25) 0.012 451 174 129 074 176 × 2 = 0 + 0.024 902 348 258 148 352;
  • 26) 0.024 902 348 258 148 352 × 2 = 0 + 0.049 804 696 516 296 704;
  • 27) 0.049 804 696 516 296 704 × 2 = 0 + 0.099 609 393 032 593 408;
  • 28) 0.099 609 393 032 593 408 × 2 = 0 + 0.199 218 786 065 186 816;
  • 29) 0.199 218 786 065 186 816 × 2 = 0 + 0.398 437 572 130 373 632;
  • 30) 0.398 437 572 130 373 632 × 2 = 0 + 0.796 875 144 260 747 264;
  • 31) 0.796 875 144 260 747 264 × 2 = 1 + 0.593 750 288 521 494 528;
  • 32) 0.593 750 288 521 494 528 × 2 = 1 + 0.187 500 577 042 989 056;
  • 33) 0.187 500 577 042 989 056 × 2 = 0 + 0.375 001 154 085 978 112;
  • 34) 0.375 001 154 085 978 112 × 2 = 0 + 0.750 002 308 171 956 224;
  • 35) 0.750 002 308 171 956 224 × 2 = 1 + 0.500 004 616 343 912 448;
  • 36) 0.500 004 616 343 912 448 × 2 = 1 + 0.000 009 232 687 824 896;
  • 37) 0.000 009 232 687 824 896 × 2 = 0 + 0.000 018 465 375 649 792;
  • 38) 0.000 018 465 375 649 792 × 2 = 0 + 0.000 036 930 751 299 584;
  • 39) 0.000 036 930 751 299 584 × 2 = 0 + 0.000 073 861 502 599 168;
  • 40) 0.000 073 861 502 599 168 × 2 = 0 + 0.000 147 723 005 198 336;
  • 41) 0.000 147 723 005 198 336 × 2 = 0 + 0.000 295 446 010 396 672;
  • 42) 0.000 295 446 010 396 672 × 2 = 0 + 0.000 590 892 020 793 344;
  • 43) 0.000 590 892 020 793 344 × 2 = 0 + 0.001 181 784 041 586 688;
  • 44) 0.001 181 784 041 586 688 × 2 = 0 + 0.002 363 568 083 173 376;
  • 45) 0.002 363 568 083 173 376 × 2 = 0 + 0.004 727 136 166 346 752;
  • 46) 0.004 727 136 166 346 752 × 2 = 0 + 0.009 454 272 332 693 504;
  • 47) 0.009 454 272 332 693 504 × 2 = 0 + 0.018 908 544 665 387 008;
  • 48) 0.018 908 544 665 387 008 × 2 = 0 + 0.037 817 089 330 774 016;
  • 49) 0.037 817 089 330 774 016 × 2 = 0 + 0.075 634 178 661 548 032;
  • 50) 0.075 634 178 661 548 032 × 2 = 0 + 0.151 268 357 323 096 064;
  • 51) 0.151 268 357 323 096 064 × 2 = 0 + 0.302 536 714 646 192 128;
  • 52) 0.302 536 714 646 192 128 × 2 = 0 + 0.605 073 429 292 384 256;
  • 53) 0.605 073 429 292 384 256 × 2 = 1 + 0.210 146 858 584 768 512;
  • 54) 0.210 146 858 584 768 512 × 2 = 0 + 0.420 293 717 169 537 024;

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:


0.000 000 000 742 147 811(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 811(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 811(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 811 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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