-0.000 000 000 742 147 676 502 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 502(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
-0.000 000 000 742 147 676 502(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:

|-0.000 000 000 742 147 676 502| = 0.000 000 000 742 147 676 502


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: 0.000 000 000 742 147 676 502.

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) 0.000 000 000 742 147 676 502 × 2 = 0 + 0.000 000 001 484 295 353 004;
  • 2) 0.000 000 001 484 295 353 004 × 2 = 0 + 0.000 000 002 968 590 706 008;
  • 3) 0.000 000 002 968 590 706 008 × 2 = 0 + 0.000 000 005 937 181 412 016;
  • 4) 0.000 000 005 937 181 412 016 × 2 = 0 + 0.000 000 011 874 362 824 032;
  • 5) 0.000 000 011 874 362 824 032 × 2 = 0 + 0.000 000 023 748 725 648 064;
  • 6) 0.000 000 023 748 725 648 064 × 2 = 0 + 0.000 000 047 497 451 296 128;
  • 7) 0.000 000 047 497 451 296 128 × 2 = 0 + 0.000 000 094 994 902 592 256;
  • 8) 0.000 000 094 994 902 592 256 × 2 = 0 + 0.000 000 189 989 805 184 512;
  • 9) 0.000 000 189 989 805 184 512 × 2 = 0 + 0.000 000 379 979 610 369 024;
  • 10) 0.000 000 379 979 610 369 024 × 2 = 0 + 0.000 000 759 959 220 738 048;
  • 11) 0.000 000 759 959 220 738 048 × 2 = 0 + 0.000 001 519 918 441 476 096;
  • 12) 0.000 001 519 918 441 476 096 × 2 = 0 + 0.000 003 039 836 882 952 192;
  • 13) 0.000 003 039 836 882 952 192 × 2 = 0 + 0.000 006 079 673 765 904 384;
  • 14) 0.000 006 079 673 765 904 384 × 2 = 0 + 0.000 012 159 347 531 808 768;
  • 15) 0.000 012 159 347 531 808 768 × 2 = 0 + 0.000 024 318 695 063 617 536;
  • 16) 0.000 024 318 695 063 617 536 × 2 = 0 + 0.000 048 637 390 127 235 072;
  • 17) 0.000 048 637 390 127 235 072 × 2 = 0 + 0.000 097 274 780 254 470 144;
  • 18) 0.000 097 274 780 254 470 144 × 2 = 0 + 0.000 194 549 560 508 940 288;
  • 19) 0.000 194 549 560 508 940 288 × 2 = 0 + 0.000 389 099 121 017 880 576;
  • 20) 0.000 389 099 121 017 880 576 × 2 = 0 + 0.000 778 198 242 035 761 152;
  • 21) 0.000 778 198 242 035 761 152 × 2 = 0 + 0.001 556 396 484 071 522 304;
  • 22) 0.001 556 396 484 071 522 304 × 2 = 0 + 0.003 112 792 968 143 044 608;
  • 23) 0.003 112 792 968 143 044 608 × 2 = 0 + 0.006 225 585 936 286 089 216;
  • 24) 0.006 225 585 936 286 089 216 × 2 = 0 + 0.012 451 171 872 572 178 432;
  • 25) 0.012 451 171 872 572 178 432 × 2 = 0 + 0.024 902 343 745 144 356 864;
  • 26) 0.024 902 343 745 144 356 864 × 2 = 0 + 0.049 804 687 490 288 713 728;
  • 27) 0.049 804 687 490 288 713 728 × 2 = 0 + 0.099 609 374 980 577 427 456;
  • 28) 0.099 609 374 980 577 427 456 × 2 = 0 + 0.199 218 749 961 154 854 912;
  • 29) 0.199 218 749 961 154 854 912 × 2 = 0 + 0.398 437 499 922 309 709 824;
  • 30) 0.398 437 499 922 309 709 824 × 2 = 0 + 0.796 874 999 844 619 419 648;
  • 31) 0.796 874 999 844 619 419 648 × 2 = 1 + 0.593 749 999 689 238 839 296;
  • 32) 0.593 749 999 689 238 839 296 × 2 = 1 + 0.187 499 999 378 477 678 592;
  • 33) 0.187 499 999 378 477 678 592 × 2 = 0 + 0.374 999 998 756 955 357 184;
  • 34) 0.374 999 998 756 955 357 184 × 2 = 0 + 0.749 999 997 513 910 714 368;
  • 35) 0.749 999 997 513 910 714 368 × 2 = 1 + 0.499 999 995 027 821 428 736;
  • 36) 0.499 999 995 027 821 428 736 × 2 = 0 + 0.999 999 990 055 642 857 472;
  • 37) 0.999 999 990 055 642 857 472 × 2 = 1 + 0.999 999 980 111 285 714 944;
  • 38) 0.999 999 980 111 285 714 944 × 2 = 1 + 0.999 999 960 222 571 429 888;
  • 39) 0.999 999 960 222 571 429 888 × 2 = 1 + 0.999 999 920 445 142 859 776;
  • 40) 0.999 999 920 445 142 859 776 × 2 = 1 + 0.999 999 840 890 285 719 552;
  • 41) 0.999 999 840 890 285 719 552 × 2 = 1 + 0.999 999 681 780 571 439 104;
  • 42) 0.999 999 681 780 571 439 104 × 2 = 1 + 0.999 999 363 561 142 878 208;
  • 43) 0.999 999 363 561 142 878 208 × 2 = 1 + 0.999 998 727 122 285 756 416;
  • 44) 0.999 998 727 122 285 756 416 × 2 = 1 + 0.999 997 454 244 571 512 832;
  • 45) 0.999 997 454 244 571 512 832 × 2 = 1 + 0.999 994 908 489 143 025 664;
  • 46) 0.999 994 908 489 143 025 664 × 2 = 1 + 0.999 989 816 978 286 051 328;
  • 47) 0.999 989 816 978 286 051 328 × 2 = 1 + 0.999 979 633 956 572 102 656;
  • 48) 0.999 979 633 956 572 102 656 × 2 = 1 + 0.999 959 267 913 144 205 312;
  • 49) 0.999 959 267 913 144 205 312 × 2 = 1 + 0.999 918 535 826 288 410 624;
  • 50) 0.999 918 535 826 288 410 624 × 2 = 1 + 0.999 837 071 652 576 821 248;
  • 51) 0.999 837 071 652 576 821 248 × 2 = 1 + 0.999 674 143 305 153 642 496;
  • 52) 0.999 674 143 305 153 642 496 × 2 = 1 + 0.999 348 286 610 307 284 992;
  • 53) 0.999 348 286 610 307 284 992 × 2 = 1 + 0.998 696 573 220 614 569 984;
  • 54) 0.998 696 573 220 614 569 984 × 2 = 1 + 0.997 393 146 441 229 139 968;

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:


0.000 000 000 742 147 676 502(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 502(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:


0.000 000 000 742 147 676 502(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 -0.000 000 000 742 147 676 502 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