-0.000 000 000 742 147 676 436 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 436(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
-0.000 000 000 742 147 676 436(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:

|-0.000 000 000 742 147 676 436| = 0.000 000 000 742 147 676 436


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: 0.000 000 000 742 147 676 436.

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) 0.000 000 000 742 147 676 436 × 2 = 0 + 0.000 000 001 484 295 352 872;
  • 2) 0.000 000 001 484 295 352 872 × 2 = 0 + 0.000 000 002 968 590 705 744;
  • 3) 0.000 000 002 968 590 705 744 × 2 = 0 + 0.000 000 005 937 181 411 488;
  • 4) 0.000 000 005 937 181 411 488 × 2 = 0 + 0.000 000 011 874 362 822 976;
  • 5) 0.000 000 011 874 362 822 976 × 2 = 0 + 0.000 000 023 748 725 645 952;
  • 6) 0.000 000 023 748 725 645 952 × 2 = 0 + 0.000 000 047 497 451 291 904;
  • 7) 0.000 000 047 497 451 291 904 × 2 = 0 + 0.000 000 094 994 902 583 808;
  • 8) 0.000 000 094 994 902 583 808 × 2 = 0 + 0.000 000 189 989 805 167 616;
  • 9) 0.000 000 189 989 805 167 616 × 2 = 0 + 0.000 000 379 979 610 335 232;
  • 10) 0.000 000 379 979 610 335 232 × 2 = 0 + 0.000 000 759 959 220 670 464;
  • 11) 0.000 000 759 959 220 670 464 × 2 = 0 + 0.000 001 519 918 441 340 928;
  • 12) 0.000 001 519 918 441 340 928 × 2 = 0 + 0.000 003 039 836 882 681 856;
  • 13) 0.000 003 039 836 882 681 856 × 2 = 0 + 0.000 006 079 673 765 363 712;
  • 14) 0.000 006 079 673 765 363 712 × 2 = 0 + 0.000 012 159 347 530 727 424;
  • 15) 0.000 012 159 347 530 727 424 × 2 = 0 + 0.000 024 318 695 061 454 848;
  • 16) 0.000 024 318 695 061 454 848 × 2 = 0 + 0.000 048 637 390 122 909 696;
  • 17) 0.000 048 637 390 122 909 696 × 2 = 0 + 0.000 097 274 780 245 819 392;
  • 18) 0.000 097 274 780 245 819 392 × 2 = 0 + 0.000 194 549 560 491 638 784;
  • 19) 0.000 194 549 560 491 638 784 × 2 = 0 + 0.000 389 099 120 983 277 568;
  • 20) 0.000 389 099 120 983 277 568 × 2 = 0 + 0.000 778 198 241 966 555 136;
  • 21) 0.000 778 198 241 966 555 136 × 2 = 0 + 0.001 556 396 483 933 110 272;
  • 22) 0.001 556 396 483 933 110 272 × 2 = 0 + 0.003 112 792 967 866 220 544;
  • 23) 0.003 112 792 967 866 220 544 × 2 = 0 + 0.006 225 585 935 732 441 088;
  • 24) 0.006 225 585 935 732 441 088 × 2 = 0 + 0.012 451 171 871 464 882 176;
  • 25) 0.012 451 171 871 464 882 176 × 2 = 0 + 0.024 902 343 742 929 764 352;
  • 26) 0.024 902 343 742 929 764 352 × 2 = 0 + 0.049 804 687 485 859 528 704;
  • 27) 0.049 804 687 485 859 528 704 × 2 = 0 + 0.099 609 374 971 719 057 408;
  • 28) 0.099 609 374 971 719 057 408 × 2 = 0 + 0.199 218 749 943 438 114 816;
  • 29) 0.199 218 749 943 438 114 816 × 2 = 0 + 0.398 437 499 886 876 229 632;
  • 30) 0.398 437 499 886 876 229 632 × 2 = 0 + 0.796 874 999 773 752 459 264;
  • 31) 0.796 874 999 773 752 459 264 × 2 = 1 + 0.593 749 999 547 504 918 528;
  • 32) 0.593 749 999 547 504 918 528 × 2 = 1 + 0.187 499 999 095 009 837 056;
  • 33) 0.187 499 999 095 009 837 056 × 2 = 0 + 0.374 999 998 190 019 674 112;
  • 34) 0.374 999 998 190 019 674 112 × 2 = 0 + 0.749 999 996 380 039 348 224;
  • 35) 0.749 999 996 380 039 348 224 × 2 = 1 + 0.499 999 992 760 078 696 448;
  • 36) 0.499 999 992 760 078 696 448 × 2 = 0 + 0.999 999 985 520 157 392 896;
  • 37) 0.999 999 985 520 157 392 896 × 2 = 1 + 0.999 999 971 040 314 785 792;
  • 38) 0.999 999 971 040 314 785 792 × 2 = 1 + 0.999 999 942 080 629 571 584;
  • 39) 0.999 999 942 080 629 571 584 × 2 = 1 + 0.999 999 884 161 259 143 168;
  • 40) 0.999 999 884 161 259 143 168 × 2 = 1 + 0.999 999 768 322 518 286 336;
  • 41) 0.999 999 768 322 518 286 336 × 2 = 1 + 0.999 999 536 645 036 572 672;
  • 42) 0.999 999 536 645 036 572 672 × 2 = 1 + 0.999 999 073 290 073 145 344;
  • 43) 0.999 999 073 290 073 145 344 × 2 = 1 + 0.999 998 146 580 146 290 688;
  • 44) 0.999 998 146 580 146 290 688 × 2 = 1 + 0.999 996 293 160 292 581 376;
  • 45) 0.999 996 293 160 292 581 376 × 2 = 1 + 0.999 992 586 320 585 162 752;
  • 46) 0.999 992 586 320 585 162 752 × 2 = 1 + 0.999 985 172 641 170 325 504;
  • 47) 0.999 985 172 641 170 325 504 × 2 = 1 + 0.999 970 345 282 340 651 008;
  • 48) 0.999 970 345 282 340 651 008 × 2 = 1 + 0.999 940 690 564 681 302 016;
  • 49) 0.999 940 690 564 681 302 016 × 2 = 1 + 0.999 881 381 129 362 604 032;
  • 50) 0.999 881 381 129 362 604 032 × 2 = 1 + 0.999 762 762 258 725 208 064;
  • 51) 0.999 762 762 258 725 208 064 × 2 = 1 + 0.999 525 524 517 450 416 128;
  • 52) 0.999 525 524 517 450 416 128 × 2 = 1 + 0.999 051 049 034 900 832 256;
  • 53) 0.999 051 049 034 900 832 256 × 2 = 1 + 0.998 102 098 069 801 664 512;
  • 54) 0.998 102 098 069 801 664 512 × 2 = 1 + 0.996 204 196 139 603 329 024;

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:


0.000 000 000 742 147 676 436(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 436(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:


0.000 000 000 742 147 676 436(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 -0.000 000 000 742 147 676 436 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