-0.000 000 000 742 147 676 634 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 634(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
-0.000 000 000 742 147 676 634(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:

|-0.000 000 000 742 147 676 634| = 0.000 000 000 742 147 676 634


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: 0.000 000 000 742 147 676 634.

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) 0.000 000 000 742 147 676 634 × 2 = 0 + 0.000 000 001 484 295 353 268;
  • 2) 0.000 000 001 484 295 353 268 × 2 = 0 + 0.000 000 002 968 590 706 536;
  • 3) 0.000 000 002 968 590 706 536 × 2 = 0 + 0.000 000 005 937 181 413 072;
  • 4) 0.000 000 005 937 181 413 072 × 2 = 0 + 0.000 000 011 874 362 826 144;
  • 5) 0.000 000 011 874 362 826 144 × 2 = 0 + 0.000 000 023 748 725 652 288;
  • 6) 0.000 000 023 748 725 652 288 × 2 = 0 + 0.000 000 047 497 451 304 576;
  • 7) 0.000 000 047 497 451 304 576 × 2 = 0 + 0.000 000 094 994 902 609 152;
  • 8) 0.000 000 094 994 902 609 152 × 2 = 0 + 0.000 000 189 989 805 218 304;
  • 9) 0.000 000 189 989 805 218 304 × 2 = 0 + 0.000 000 379 979 610 436 608;
  • 10) 0.000 000 379 979 610 436 608 × 2 = 0 + 0.000 000 759 959 220 873 216;
  • 11) 0.000 000 759 959 220 873 216 × 2 = 0 + 0.000 001 519 918 441 746 432;
  • 12) 0.000 001 519 918 441 746 432 × 2 = 0 + 0.000 003 039 836 883 492 864;
  • 13) 0.000 003 039 836 883 492 864 × 2 = 0 + 0.000 006 079 673 766 985 728;
  • 14) 0.000 006 079 673 766 985 728 × 2 = 0 + 0.000 012 159 347 533 971 456;
  • 15) 0.000 012 159 347 533 971 456 × 2 = 0 + 0.000 024 318 695 067 942 912;
  • 16) 0.000 024 318 695 067 942 912 × 2 = 0 + 0.000 048 637 390 135 885 824;
  • 17) 0.000 048 637 390 135 885 824 × 2 = 0 + 0.000 097 274 780 271 771 648;
  • 18) 0.000 097 274 780 271 771 648 × 2 = 0 + 0.000 194 549 560 543 543 296;
  • 19) 0.000 194 549 560 543 543 296 × 2 = 0 + 0.000 389 099 121 087 086 592;
  • 20) 0.000 389 099 121 087 086 592 × 2 = 0 + 0.000 778 198 242 174 173 184;
  • 21) 0.000 778 198 242 174 173 184 × 2 = 0 + 0.001 556 396 484 348 346 368;
  • 22) 0.001 556 396 484 348 346 368 × 2 = 0 + 0.003 112 792 968 696 692 736;
  • 23) 0.003 112 792 968 696 692 736 × 2 = 0 + 0.006 225 585 937 393 385 472;
  • 24) 0.006 225 585 937 393 385 472 × 2 = 0 + 0.012 451 171 874 786 770 944;
  • 25) 0.012 451 171 874 786 770 944 × 2 = 0 + 0.024 902 343 749 573 541 888;
  • 26) 0.024 902 343 749 573 541 888 × 2 = 0 + 0.049 804 687 499 147 083 776;
  • 27) 0.049 804 687 499 147 083 776 × 2 = 0 + 0.099 609 374 998 294 167 552;
  • 28) 0.099 609 374 998 294 167 552 × 2 = 0 + 0.199 218 749 996 588 335 104;
  • 29) 0.199 218 749 996 588 335 104 × 2 = 0 + 0.398 437 499 993 176 670 208;
  • 30) 0.398 437 499 993 176 670 208 × 2 = 0 + 0.796 874 999 986 353 340 416;
  • 31) 0.796 874 999 986 353 340 416 × 2 = 1 + 0.593 749 999 972 706 680 832;
  • 32) 0.593 749 999 972 706 680 832 × 2 = 1 + 0.187 499 999 945 413 361 664;
  • 33) 0.187 499 999 945 413 361 664 × 2 = 0 + 0.374 999 999 890 826 723 328;
  • 34) 0.374 999 999 890 826 723 328 × 2 = 0 + 0.749 999 999 781 653 446 656;
  • 35) 0.749 999 999 781 653 446 656 × 2 = 1 + 0.499 999 999 563 306 893 312;
  • 36) 0.499 999 999 563 306 893 312 × 2 = 0 + 0.999 999 999 126 613 786 624;
  • 37) 0.999 999 999 126 613 786 624 × 2 = 1 + 0.999 999 998 253 227 573 248;
  • 38) 0.999 999 998 253 227 573 248 × 2 = 1 + 0.999 999 996 506 455 146 496;
  • 39) 0.999 999 996 506 455 146 496 × 2 = 1 + 0.999 999 993 012 910 292 992;
  • 40) 0.999 999 993 012 910 292 992 × 2 = 1 + 0.999 999 986 025 820 585 984;
  • 41) 0.999 999 986 025 820 585 984 × 2 = 1 + 0.999 999 972 051 641 171 968;
  • 42) 0.999 999 972 051 641 171 968 × 2 = 1 + 0.999 999 944 103 282 343 936;
  • 43) 0.999 999 944 103 282 343 936 × 2 = 1 + 0.999 999 888 206 564 687 872;
  • 44) 0.999 999 888 206 564 687 872 × 2 = 1 + 0.999 999 776 413 129 375 744;
  • 45) 0.999 999 776 413 129 375 744 × 2 = 1 + 0.999 999 552 826 258 751 488;
  • 46) 0.999 999 552 826 258 751 488 × 2 = 1 + 0.999 999 105 652 517 502 976;
  • 47) 0.999 999 105 652 517 502 976 × 2 = 1 + 0.999 998 211 305 035 005 952;
  • 48) 0.999 998 211 305 035 005 952 × 2 = 1 + 0.999 996 422 610 070 011 904;
  • 49) 0.999 996 422 610 070 011 904 × 2 = 1 + 0.999 992 845 220 140 023 808;
  • 50) 0.999 992 845 220 140 023 808 × 2 = 1 + 0.999 985 690 440 280 047 616;
  • 51) 0.999 985 690 440 280 047 616 × 2 = 1 + 0.999 971 380 880 560 095 232;
  • 52) 0.999 971 380 880 560 095 232 × 2 = 1 + 0.999 942 761 761 120 190 464;
  • 53) 0.999 942 761 761 120 190 464 × 2 = 1 + 0.999 885 523 522 240 380 928;
  • 54) 0.999 885 523 522 240 380 928 × 2 = 1 + 0.999 771 047 044 480 761 856;

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:


0.000 000 000 742 147 676 634(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 634(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:


0.000 000 000 742 147 676 634(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 -0.000 000 000 742 147 676 634 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