-0.000 000 000 742 147 553 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 553(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
-0.000 000 000 742 147 553(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:

|-0.000 000 000 742 147 553| = 0.000 000 000 742 147 553


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: 0.000 000 000 742 147 553.

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) 0.000 000 000 742 147 553 × 2 = 0 + 0.000 000 001 484 295 106;
  • 2) 0.000 000 001 484 295 106 × 2 = 0 + 0.000 000 002 968 590 212;
  • 3) 0.000 000 002 968 590 212 × 2 = 0 + 0.000 000 005 937 180 424;
  • 4) 0.000 000 005 937 180 424 × 2 = 0 + 0.000 000 011 874 360 848;
  • 5) 0.000 000 011 874 360 848 × 2 = 0 + 0.000 000 023 748 721 696;
  • 6) 0.000 000 023 748 721 696 × 2 = 0 + 0.000 000 047 497 443 392;
  • 7) 0.000 000 047 497 443 392 × 2 = 0 + 0.000 000 094 994 886 784;
  • 8) 0.000 000 094 994 886 784 × 2 = 0 + 0.000 000 189 989 773 568;
  • 9) 0.000 000 189 989 773 568 × 2 = 0 + 0.000 000 379 979 547 136;
  • 10) 0.000 000 379 979 547 136 × 2 = 0 + 0.000 000 759 959 094 272;
  • 11) 0.000 000 759 959 094 272 × 2 = 0 + 0.000 001 519 918 188 544;
  • 12) 0.000 001 519 918 188 544 × 2 = 0 + 0.000 003 039 836 377 088;
  • 13) 0.000 003 039 836 377 088 × 2 = 0 + 0.000 006 079 672 754 176;
  • 14) 0.000 006 079 672 754 176 × 2 = 0 + 0.000 012 159 345 508 352;
  • 15) 0.000 012 159 345 508 352 × 2 = 0 + 0.000 024 318 691 016 704;
  • 16) 0.000 024 318 691 016 704 × 2 = 0 + 0.000 048 637 382 033 408;
  • 17) 0.000 048 637 382 033 408 × 2 = 0 + 0.000 097 274 764 066 816;
  • 18) 0.000 097 274 764 066 816 × 2 = 0 + 0.000 194 549 528 133 632;
  • 19) 0.000 194 549 528 133 632 × 2 = 0 + 0.000 389 099 056 267 264;
  • 20) 0.000 389 099 056 267 264 × 2 = 0 + 0.000 778 198 112 534 528;
  • 21) 0.000 778 198 112 534 528 × 2 = 0 + 0.001 556 396 225 069 056;
  • 22) 0.001 556 396 225 069 056 × 2 = 0 + 0.003 112 792 450 138 112;
  • 23) 0.003 112 792 450 138 112 × 2 = 0 + 0.006 225 584 900 276 224;
  • 24) 0.006 225 584 900 276 224 × 2 = 0 + 0.012 451 169 800 552 448;
  • 25) 0.012 451 169 800 552 448 × 2 = 0 + 0.024 902 339 601 104 896;
  • 26) 0.024 902 339 601 104 896 × 2 = 0 + 0.049 804 679 202 209 792;
  • 27) 0.049 804 679 202 209 792 × 2 = 0 + 0.099 609 358 404 419 584;
  • 28) 0.099 609 358 404 419 584 × 2 = 0 + 0.199 218 716 808 839 168;
  • 29) 0.199 218 716 808 839 168 × 2 = 0 + 0.398 437 433 617 678 336;
  • 30) 0.398 437 433 617 678 336 × 2 = 0 + 0.796 874 867 235 356 672;
  • 31) 0.796 874 867 235 356 672 × 2 = 1 + 0.593 749 734 470 713 344;
  • 32) 0.593 749 734 470 713 344 × 2 = 1 + 0.187 499 468 941 426 688;
  • 33) 0.187 499 468 941 426 688 × 2 = 0 + 0.374 998 937 882 853 376;
  • 34) 0.374 998 937 882 853 376 × 2 = 0 + 0.749 997 875 765 706 752;
  • 35) 0.749 997 875 765 706 752 × 2 = 1 + 0.499 995 751 531 413 504;
  • 36) 0.499 995 751 531 413 504 × 2 = 0 + 0.999 991 503 062 827 008;
  • 37) 0.999 991 503 062 827 008 × 2 = 1 + 0.999 983 006 125 654 016;
  • 38) 0.999 983 006 125 654 016 × 2 = 1 + 0.999 966 012 251 308 032;
  • 39) 0.999 966 012 251 308 032 × 2 = 1 + 0.999 932 024 502 616 064;
  • 40) 0.999 932 024 502 616 064 × 2 = 1 + 0.999 864 049 005 232 128;
  • 41) 0.999 864 049 005 232 128 × 2 = 1 + 0.999 728 098 010 464 256;
  • 42) 0.999 728 098 010 464 256 × 2 = 1 + 0.999 456 196 020 928 512;
  • 43) 0.999 456 196 020 928 512 × 2 = 1 + 0.998 912 392 041 857 024;
  • 44) 0.998 912 392 041 857 024 × 2 = 1 + 0.997 824 784 083 714 048;
  • 45) 0.997 824 784 083 714 048 × 2 = 1 + 0.995 649 568 167 428 096;
  • 46) 0.995 649 568 167 428 096 × 2 = 1 + 0.991 299 136 334 856 192;
  • 47) 0.991 299 136 334 856 192 × 2 = 1 + 0.982 598 272 669 712 384;
  • 48) 0.982 598 272 669 712 384 × 2 = 1 + 0.965 196 545 339 424 768;
  • 49) 0.965 196 545 339 424 768 × 2 = 1 + 0.930 393 090 678 849 536;
  • 50) 0.930 393 090 678 849 536 × 2 = 1 + 0.860 786 181 357 699 072;
  • 51) 0.860 786 181 357 699 072 × 2 = 1 + 0.721 572 362 715 398 144;
  • 52) 0.721 572 362 715 398 144 × 2 = 1 + 0.443 144 725 430 796 288;
  • 53) 0.443 144 725 430 796 288 × 2 = 0 + 0.886 289 450 861 592 576;
  • 54) 0.886 289 450 861 592 576 × 2 = 1 + 0.772 578 901 723 185 152;

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:


0.000 000 000 742 147 553(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 553(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:


0.000 000 000 742 147 553(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 -0.000 000 000 742 147 553 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