-0.000 000 000 742 147 676 438 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 438(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
-0.000 000 000 742 147 676 438(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:

|-0.000 000 000 742 147 676 438| = 0.000 000 000 742 147 676 438


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: 0.000 000 000 742 147 676 438.

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) 0.000 000 000 742 147 676 438 × 2 = 0 + 0.000 000 001 484 295 352 876;
  • 2) 0.000 000 001 484 295 352 876 × 2 = 0 + 0.000 000 002 968 590 705 752;
  • 3) 0.000 000 002 968 590 705 752 × 2 = 0 + 0.000 000 005 937 181 411 504;
  • 4) 0.000 000 005 937 181 411 504 × 2 = 0 + 0.000 000 011 874 362 823 008;
  • 5) 0.000 000 011 874 362 823 008 × 2 = 0 + 0.000 000 023 748 725 646 016;
  • 6) 0.000 000 023 748 725 646 016 × 2 = 0 + 0.000 000 047 497 451 292 032;
  • 7) 0.000 000 047 497 451 292 032 × 2 = 0 + 0.000 000 094 994 902 584 064;
  • 8) 0.000 000 094 994 902 584 064 × 2 = 0 + 0.000 000 189 989 805 168 128;
  • 9) 0.000 000 189 989 805 168 128 × 2 = 0 + 0.000 000 379 979 610 336 256;
  • 10) 0.000 000 379 979 610 336 256 × 2 = 0 + 0.000 000 759 959 220 672 512;
  • 11) 0.000 000 759 959 220 672 512 × 2 = 0 + 0.000 001 519 918 441 345 024;
  • 12) 0.000 001 519 918 441 345 024 × 2 = 0 + 0.000 003 039 836 882 690 048;
  • 13) 0.000 003 039 836 882 690 048 × 2 = 0 + 0.000 006 079 673 765 380 096;
  • 14) 0.000 006 079 673 765 380 096 × 2 = 0 + 0.000 012 159 347 530 760 192;
  • 15) 0.000 012 159 347 530 760 192 × 2 = 0 + 0.000 024 318 695 061 520 384;
  • 16) 0.000 024 318 695 061 520 384 × 2 = 0 + 0.000 048 637 390 123 040 768;
  • 17) 0.000 048 637 390 123 040 768 × 2 = 0 + 0.000 097 274 780 246 081 536;
  • 18) 0.000 097 274 780 246 081 536 × 2 = 0 + 0.000 194 549 560 492 163 072;
  • 19) 0.000 194 549 560 492 163 072 × 2 = 0 + 0.000 389 099 120 984 326 144;
  • 20) 0.000 389 099 120 984 326 144 × 2 = 0 + 0.000 778 198 241 968 652 288;
  • 21) 0.000 778 198 241 968 652 288 × 2 = 0 + 0.001 556 396 483 937 304 576;
  • 22) 0.001 556 396 483 937 304 576 × 2 = 0 + 0.003 112 792 967 874 609 152;
  • 23) 0.003 112 792 967 874 609 152 × 2 = 0 + 0.006 225 585 935 749 218 304;
  • 24) 0.006 225 585 935 749 218 304 × 2 = 0 + 0.012 451 171 871 498 436 608;
  • 25) 0.012 451 171 871 498 436 608 × 2 = 0 + 0.024 902 343 742 996 873 216;
  • 26) 0.024 902 343 742 996 873 216 × 2 = 0 + 0.049 804 687 485 993 746 432;
  • 27) 0.049 804 687 485 993 746 432 × 2 = 0 + 0.099 609 374 971 987 492 864;
  • 28) 0.099 609 374 971 987 492 864 × 2 = 0 + 0.199 218 749 943 974 985 728;
  • 29) 0.199 218 749 943 974 985 728 × 2 = 0 + 0.398 437 499 887 949 971 456;
  • 30) 0.398 437 499 887 949 971 456 × 2 = 0 + 0.796 874 999 775 899 942 912;
  • 31) 0.796 874 999 775 899 942 912 × 2 = 1 + 0.593 749 999 551 799 885 824;
  • 32) 0.593 749 999 551 799 885 824 × 2 = 1 + 0.187 499 999 103 599 771 648;
  • 33) 0.187 499 999 103 599 771 648 × 2 = 0 + 0.374 999 998 207 199 543 296;
  • 34) 0.374 999 998 207 199 543 296 × 2 = 0 + 0.749 999 996 414 399 086 592;
  • 35) 0.749 999 996 414 399 086 592 × 2 = 1 + 0.499 999 992 828 798 173 184;
  • 36) 0.499 999 992 828 798 173 184 × 2 = 0 + 0.999 999 985 657 596 346 368;
  • 37) 0.999 999 985 657 596 346 368 × 2 = 1 + 0.999 999 971 315 192 692 736;
  • 38) 0.999 999 971 315 192 692 736 × 2 = 1 + 0.999 999 942 630 385 385 472;
  • 39) 0.999 999 942 630 385 385 472 × 2 = 1 + 0.999 999 885 260 770 770 944;
  • 40) 0.999 999 885 260 770 770 944 × 2 = 1 + 0.999 999 770 521 541 541 888;
  • 41) 0.999 999 770 521 541 541 888 × 2 = 1 + 0.999 999 541 043 083 083 776;
  • 42) 0.999 999 541 043 083 083 776 × 2 = 1 + 0.999 999 082 086 166 167 552;
  • 43) 0.999 999 082 086 166 167 552 × 2 = 1 + 0.999 998 164 172 332 335 104;
  • 44) 0.999 998 164 172 332 335 104 × 2 = 1 + 0.999 996 328 344 664 670 208;
  • 45) 0.999 996 328 344 664 670 208 × 2 = 1 + 0.999 992 656 689 329 340 416;
  • 46) 0.999 992 656 689 329 340 416 × 2 = 1 + 0.999 985 313 378 658 680 832;
  • 47) 0.999 985 313 378 658 680 832 × 2 = 1 + 0.999 970 626 757 317 361 664;
  • 48) 0.999 970 626 757 317 361 664 × 2 = 1 + 0.999 941 253 514 634 723 328;
  • 49) 0.999 941 253 514 634 723 328 × 2 = 1 + 0.999 882 507 029 269 446 656;
  • 50) 0.999 882 507 029 269 446 656 × 2 = 1 + 0.999 765 014 058 538 893 312;
  • 51) 0.999 765 014 058 538 893 312 × 2 = 1 + 0.999 530 028 117 077 786 624;
  • 52) 0.999 530 028 117 077 786 624 × 2 = 1 + 0.999 060 056 234 155 573 248;
  • 53) 0.999 060 056 234 155 573 248 × 2 = 1 + 0.998 120 112 468 311 146 496;
  • 54) 0.998 120 112 468 311 146 496 × 2 = 1 + 0.996 240 224 936 622 292 992;

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:


0.000 000 000 742 147 676 438(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 438(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:


0.000 000 000 742 147 676 438(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 -0.000 000 000 742 147 676 438 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