-0.000 000 000 742 147 818 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 818(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
-0.000 000 000 742 147 818(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:

|-0.000 000 000 742 147 818| = 0.000 000 000 742 147 818


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: 0.000 000 000 742 147 818.

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) 0.000 000 000 742 147 818 × 2 = 0 + 0.000 000 001 484 295 636;
  • 2) 0.000 000 001 484 295 636 × 2 = 0 + 0.000 000 002 968 591 272;
  • 3) 0.000 000 002 968 591 272 × 2 = 0 + 0.000 000 005 937 182 544;
  • 4) 0.000 000 005 937 182 544 × 2 = 0 + 0.000 000 011 874 365 088;
  • 5) 0.000 000 011 874 365 088 × 2 = 0 + 0.000 000 023 748 730 176;
  • 6) 0.000 000 023 748 730 176 × 2 = 0 + 0.000 000 047 497 460 352;
  • 7) 0.000 000 047 497 460 352 × 2 = 0 + 0.000 000 094 994 920 704;
  • 8) 0.000 000 094 994 920 704 × 2 = 0 + 0.000 000 189 989 841 408;
  • 9) 0.000 000 189 989 841 408 × 2 = 0 + 0.000 000 379 979 682 816;
  • 10) 0.000 000 379 979 682 816 × 2 = 0 + 0.000 000 759 959 365 632;
  • 11) 0.000 000 759 959 365 632 × 2 = 0 + 0.000 001 519 918 731 264;
  • 12) 0.000 001 519 918 731 264 × 2 = 0 + 0.000 003 039 837 462 528;
  • 13) 0.000 003 039 837 462 528 × 2 = 0 + 0.000 006 079 674 925 056;
  • 14) 0.000 006 079 674 925 056 × 2 = 0 + 0.000 012 159 349 850 112;
  • 15) 0.000 012 159 349 850 112 × 2 = 0 + 0.000 024 318 699 700 224;
  • 16) 0.000 024 318 699 700 224 × 2 = 0 + 0.000 048 637 399 400 448;
  • 17) 0.000 048 637 399 400 448 × 2 = 0 + 0.000 097 274 798 800 896;
  • 18) 0.000 097 274 798 800 896 × 2 = 0 + 0.000 194 549 597 601 792;
  • 19) 0.000 194 549 597 601 792 × 2 = 0 + 0.000 389 099 195 203 584;
  • 20) 0.000 389 099 195 203 584 × 2 = 0 + 0.000 778 198 390 407 168;
  • 21) 0.000 778 198 390 407 168 × 2 = 0 + 0.001 556 396 780 814 336;
  • 22) 0.001 556 396 780 814 336 × 2 = 0 + 0.003 112 793 561 628 672;
  • 23) 0.003 112 793 561 628 672 × 2 = 0 + 0.006 225 587 123 257 344;
  • 24) 0.006 225 587 123 257 344 × 2 = 0 + 0.012 451 174 246 514 688;
  • 25) 0.012 451 174 246 514 688 × 2 = 0 + 0.024 902 348 493 029 376;
  • 26) 0.024 902 348 493 029 376 × 2 = 0 + 0.049 804 696 986 058 752;
  • 27) 0.049 804 696 986 058 752 × 2 = 0 + 0.099 609 393 972 117 504;
  • 28) 0.099 609 393 972 117 504 × 2 = 0 + 0.199 218 787 944 235 008;
  • 29) 0.199 218 787 944 235 008 × 2 = 0 + 0.398 437 575 888 470 016;
  • 30) 0.398 437 575 888 470 016 × 2 = 0 + 0.796 875 151 776 940 032;
  • 31) 0.796 875 151 776 940 032 × 2 = 1 + 0.593 750 303 553 880 064;
  • 32) 0.593 750 303 553 880 064 × 2 = 1 + 0.187 500 607 107 760 128;
  • 33) 0.187 500 607 107 760 128 × 2 = 0 + 0.375 001 214 215 520 256;
  • 34) 0.375 001 214 215 520 256 × 2 = 0 + 0.750 002 428 431 040 512;
  • 35) 0.750 002 428 431 040 512 × 2 = 1 + 0.500 004 856 862 081 024;
  • 36) 0.500 004 856 862 081 024 × 2 = 1 + 0.000 009 713 724 162 048;
  • 37) 0.000 009 713 724 162 048 × 2 = 0 + 0.000 019 427 448 324 096;
  • 38) 0.000 019 427 448 324 096 × 2 = 0 + 0.000 038 854 896 648 192;
  • 39) 0.000 038 854 896 648 192 × 2 = 0 + 0.000 077 709 793 296 384;
  • 40) 0.000 077 709 793 296 384 × 2 = 0 + 0.000 155 419 586 592 768;
  • 41) 0.000 155 419 586 592 768 × 2 = 0 + 0.000 310 839 173 185 536;
  • 42) 0.000 310 839 173 185 536 × 2 = 0 + 0.000 621 678 346 371 072;
  • 43) 0.000 621 678 346 371 072 × 2 = 0 + 0.001 243 356 692 742 144;
  • 44) 0.001 243 356 692 742 144 × 2 = 0 + 0.002 486 713 385 484 288;
  • 45) 0.002 486 713 385 484 288 × 2 = 0 + 0.004 973 426 770 968 576;
  • 46) 0.004 973 426 770 968 576 × 2 = 0 + 0.009 946 853 541 937 152;
  • 47) 0.009 946 853 541 937 152 × 2 = 0 + 0.019 893 707 083 874 304;
  • 48) 0.019 893 707 083 874 304 × 2 = 0 + 0.039 787 414 167 748 608;
  • 49) 0.039 787 414 167 748 608 × 2 = 0 + 0.079 574 828 335 497 216;
  • 50) 0.079 574 828 335 497 216 × 2 = 0 + 0.159 149 656 670 994 432;
  • 51) 0.159 149 656 670 994 432 × 2 = 0 + 0.318 299 313 341 988 864;
  • 52) 0.318 299 313 341 988 864 × 2 = 0 + 0.636 598 626 683 977 728;
  • 53) 0.636 598 626 683 977 728 × 2 = 1 + 0.273 197 253 367 955 456;
  • 54) 0.273 197 253 367 955 456 × 2 = 0 + 0.546 394 506 735 910 912;

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:


0.000 000 000 742 147 818(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 818(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:


0.000 000 000 742 147 818(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 -0.000 000 000 742 147 818 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