-0.000 000 000 742 147 676 652 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 652(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
-0.000 000 000 742 147 676 652(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:

|-0.000 000 000 742 147 676 652| = 0.000 000 000 742 147 676 652


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: 0.000 000 000 742 147 676 652.

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) 0.000 000 000 742 147 676 652 × 2 = 0 + 0.000 000 001 484 295 353 304;
  • 2) 0.000 000 001 484 295 353 304 × 2 = 0 + 0.000 000 002 968 590 706 608;
  • 3) 0.000 000 002 968 590 706 608 × 2 = 0 + 0.000 000 005 937 181 413 216;
  • 4) 0.000 000 005 937 181 413 216 × 2 = 0 + 0.000 000 011 874 362 826 432;
  • 5) 0.000 000 011 874 362 826 432 × 2 = 0 + 0.000 000 023 748 725 652 864;
  • 6) 0.000 000 023 748 725 652 864 × 2 = 0 + 0.000 000 047 497 451 305 728;
  • 7) 0.000 000 047 497 451 305 728 × 2 = 0 + 0.000 000 094 994 902 611 456;
  • 8) 0.000 000 094 994 902 611 456 × 2 = 0 + 0.000 000 189 989 805 222 912;
  • 9) 0.000 000 189 989 805 222 912 × 2 = 0 + 0.000 000 379 979 610 445 824;
  • 10) 0.000 000 379 979 610 445 824 × 2 = 0 + 0.000 000 759 959 220 891 648;
  • 11) 0.000 000 759 959 220 891 648 × 2 = 0 + 0.000 001 519 918 441 783 296;
  • 12) 0.000 001 519 918 441 783 296 × 2 = 0 + 0.000 003 039 836 883 566 592;
  • 13) 0.000 003 039 836 883 566 592 × 2 = 0 + 0.000 006 079 673 767 133 184;
  • 14) 0.000 006 079 673 767 133 184 × 2 = 0 + 0.000 012 159 347 534 266 368;
  • 15) 0.000 012 159 347 534 266 368 × 2 = 0 + 0.000 024 318 695 068 532 736;
  • 16) 0.000 024 318 695 068 532 736 × 2 = 0 + 0.000 048 637 390 137 065 472;
  • 17) 0.000 048 637 390 137 065 472 × 2 = 0 + 0.000 097 274 780 274 130 944;
  • 18) 0.000 097 274 780 274 130 944 × 2 = 0 + 0.000 194 549 560 548 261 888;
  • 19) 0.000 194 549 560 548 261 888 × 2 = 0 + 0.000 389 099 121 096 523 776;
  • 20) 0.000 389 099 121 096 523 776 × 2 = 0 + 0.000 778 198 242 193 047 552;
  • 21) 0.000 778 198 242 193 047 552 × 2 = 0 + 0.001 556 396 484 386 095 104;
  • 22) 0.001 556 396 484 386 095 104 × 2 = 0 + 0.003 112 792 968 772 190 208;
  • 23) 0.003 112 792 968 772 190 208 × 2 = 0 + 0.006 225 585 937 544 380 416;
  • 24) 0.006 225 585 937 544 380 416 × 2 = 0 + 0.012 451 171 875 088 760 832;
  • 25) 0.012 451 171 875 088 760 832 × 2 = 0 + 0.024 902 343 750 177 521 664;
  • 26) 0.024 902 343 750 177 521 664 × 2 = 0 + 0.049 804 687 500 355 043 328;
  • 27) 0.049 804 687 500 355 043 328 × 2 = 0 + 0.099 609 375 000 710 086 656;
  • 28) 0.099 609 375 000 710 086 656 × 2 = 0 + 0.199 218 750 001 420 173 312;
  • 29) 0.199 218 750 001 420 173 312 × 2 = 0 + 0.398 437 500 002 840 346 624;
  • 30) 0.398 437 500 002 840 346 624 × 2 = 0 + 0.796 875 000 005 680 693 248;
  • 31) 0.796 875 000 005 680 693 248 × 2 = 1 + 0.593 750 000 011 361 386 496;
  • 32) 0.593 750 000 011 361 386 496 × 2 = 1 + 0.187 500 000 022 722 772 992;
  • 33) 0.187 500 000 022 722 772 992 × 2 = 0 + 0.375 000 000 045 445 545 984;
  • 34) 0.375 000 000 045 445 545 984 × 2 = 0 + 0.750 000 000 090 891 091 968;
  • 35) 0.750 000 000 090 891 091 968 × 2 = 1 + 0.500 000 000 181 782 183 936;
  • 36) 0.500 000 000 181 782 183 936 × 2 = 1 + 0.000 000 000 363 564 367 872;
  • 37) 0.000 000 000 363 564 367 872 × 2 = 0 + 0.000 000 000 727 128 735 744;
  • 38) 0.000 000 000 727 128 735 744 × 2 = 0 + 0.000 000 001 454 257 471 488;
  • 39) 0.000 000 001 454 257 471 488 × 2 = 0 + 0.000 000 002 908 514 942 976;
  • 40) 0.000 000 002 908 514 942 976 × 2 = 0 + 0.000 000 005 817 029 885 952;
  • 41) 0.000 000 005 817 029 885 952 × 2 = 0 + 0.000 000 011 634 059 771 904;
  • 42) 0.000 000 011 634 059 771 904 × 2 = 0 + 0.000 000 023 268 119 543 808;
  • 43) 0.000 000 023 268 119 543 808 × 2 = 0 + 0.000 000 046 536 239 087 616;
  • 44) 0.000 000 046 536 239 087 616 × 2 = 0 + 0.000 000 093 072 478 175 232;
  • 45) 0.000 000 093 072 478 175 232 × 2 = 0 + 0.000 000 186 144 956 350 464;
  • 46) 0.000 000 186 144 956 350 464 × 2 = 0 + 0.000 000 372 289 912 700 928;
  • 47) 0.000 000 372 289 912 700 928 × 2 = 0 + 0.000 000 744 579 825 401 856;
  • 48) 0.000 000 744 579 825 401 856 × 2 = 0 + 0.000 001 489 159 650 803 712;
  • 49) 0.000 001 489 159 650 803 712 × 2 = 0 + 0.000 002 978 319 301 607 424;
  • 50) 0.000 002 978 319 301 607 424 × 2 = 0 + 0.000 005 956 638 603 214 848;
  • 51) 0.000 005 956 638 603 214 848 × 2 = 0 + 0.000 011 913 277 206 429 696;
  • 52) 0.000 011 913 277 206 429 696 × 2 = 0 + 0.000 023 826 554 412 859 392;
  • 53) 0.000 023 826 554 412 859 392 × 2 = 0 + 0.000 047 653 108 825 718 784;
  • 54) 0.000 047 653 108 825 718 784 × 2 = 0 + 0.000 095 306 217 651 437 568;

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:


0.000 000 000 742 147 676 652(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 652(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 652(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 676 652 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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