-0.000 000 000 742 147 676 637 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 637 3(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
-0.000 000 000 742 147 676 637 3(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:

|-0.000 000 000 742 147 676 637 3| = 0.000 000 000 742 147 676 637 3


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: 0.000 000 000 742 147 676 637 3.

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) 0.000 000 000 742 147 676 637 3 × 2 = 0 + 0.000 000 001 484 295 353 274 6;
  • 2) 0.000 000 001 484 295 353 274 6 × 2 = 0 + 0.000 000 002 968 590 706 549 2;
  • 3) 0.000 000 002 968 590 706 549 2 × 2 = 0 + 0.000 000 005 937 181 413 098 4;
  • 4) 0.000 000 005 937 181 413 098 4 × 2 = 0 + 0.000 000 011 874 362 826 196 8;
  • 5) 0.000 000 011 874 362 826 196 8 × 2 = 0 + 0.000 000 023 748 725 652 393 6;
  • 6) 0.000 000 023 748 725 652 393 6 × 2 = 0 + 0.000 000 047 497 451 304 787 2;
  • 7) 0.000 000 047 497 451 304 787 2 × 2 = 0 + 0.000 000 094 994 902 609 574 4;
  • 8) 0.000 000 094 994 902 609 574 4 × 2 = 0 + 0.000 000 189 989 805 219 148 8;
  • 9) 0.000 000 189 989 805 219 148 8 × 2 = 0 + 0.000 000 379 979 610 438 297 6;
  • 10) 0.000 000 379 979 610 438 297 6 × 2 = 0 + 0.000 000 759 959 220 876 595 2;
  • 11) 0.000 000 759 959 220 876 595 2 × 2 = 0 + 0.000 001 519 918 441 753 190 4;
  • 12) 0.000 001 519 918 441 753 190 4 × 2 = 0 + 0.000 003 039 836 883 506 380 8;
  • 13) 0.000 003 039 836 883 506 380 8 × 2 = 0 + 0.000 006 079 673 767 012 761 6;
  • 14) 0.000 006 079 673 767 012 761 6 × 2 = 0 + 0.000 012 159 347 534 025 523 2;
  • 15) 0.000 012 159 347 534 025 523 2 × 2 = 0 + 0.000 024 318 695 068 051 046 4;
  • 16) 0.000 024 318 695 068 051 046 4 × 2 = 0 + 0.000 048 637 390 136 102 092 8;
  • 17) 0.000 048 637 390 136 102 092 8 × 2 = 0 + 0.000 097 274 780 272 204 185 6;
  • 18) 0.000 097 274 780 272 204 185 6 × 2 = 0 + 0.000 194 549 560 544 408 371 2;
  • 19) 0.000 194 549 560 544 408 371 2 × 2 = 0 + 0.000 389 099 121 088 816 742 4;
  • 20) 0.000 389 099 121 088 816 742 4 × 2 = 0 + 0.000 778 198 242 177 633 484 8;
  • 21) 0.000 778 198 242 177 633 484 8 × 2 = 0 + 0.001 556 396 484 355 266 969 6;
  • 22) 0.001 556 396 484 355 266 969 6 × 2 = 0 + 0.003 112 792 968 710 533 939 2;
  • 23) 0.003 112 792 968 710 533 939 2 × 2 = 0 + 0.006 225 585 937 421 067 878 4;
  • 24) 0.006 225 585 937 421 067 878 4 × 2 = 0 + 0.012 451 171 874 842 135 756 8;
  • 25) 0.012 451 171 874 842 135 756 8 × 2 = 0 + 0.024 902 343 749 684 271 513 6;
  • 26) 0.024 902 343 749 684 271 513 6 × 2 = 0 + 0.049 804 687 499 368 543 027 2;
  • 27) 0.049 804 687 499 368 543 027 2 × 2 = 0 + 0.099 609 374 998 737 086 054 4;
  • 28) 0.099 609 374 998 737 086 054 4 × 2 = 0 + 0.199 218 749 997 474 172 108 8;
  • 29) 0.199 218 749 997 474 172 108 8 × 2 = 0 + 0.398 437 499 994 948 344 217 6;
  • 30) 0.398 437 499 994 948 344 217 6 × 2 = 0 + 0.796 874 999 989 896 688 435 2;
  • 31) 0.796 874 999 989 896 688 435 2 × 2 = 1 + 0.593 749 999 979 793 376 870 4;
  • 32) 0.593 749 999 979 793 376 870 4 × 2 = 1 + 0.187 499 999 959 586 753 740 8;
  • 33) 0.187 499 999 959 586 753 740 8 × 2 = 0 + 0.374 999 999 919 173 507 481 6;
  • 34) 0.374 999 999 919 173 507 481 6 × 2 = 0 + 0.749 999 999 838 347 014 963 2;
  • 35) 0.749 999 999 838 347 014 963 2 × 2 = 1 + 0.499 999 999 676 694 029 926 4;
  • 36) 0.499 999 999 676 694 029 926 4 × 2 = 0 + 0.999 999 999 353 388 059 852 8;
  • 37) 0.999 999 999 353 388 059 852 8 × 2 = 1 + 0.999 999 998 706 776 119 705 6;
  • 38) 0.999 999 998 706 776 119 705 6 × 2 = 1 + 0.999 999 997 413 552 239 411 2;
  • 39) 0.999 999 997 413 552 239 411 2 × 2 = 1 + 0.999 999 994 827 104 478 822 4;
  • 40) 0.999 999 994 827 104 478 822 4 × 2 = 1 + 0.999 999 989 654 208 957 644 8;
  • 41) 0.999 999 989 654 208 957 644 8 × 2 = 1 + 0.999 999 979 308 417 915 289 6;
  • 42) 0.999 999 979 308 417 915 289 6 × 2 = 1 + 0.999 999 958 616 835 830 579 2;
  • 43) 0.999 999 958 616 835 830 579 2 × 2 = 1 + 0.999 999 917 233 671 661 158 4;
  • 44) 0.999 999 917 233 671 661 158 4 × 2 = 1 + 0.999 999 834 467 343 322 316 8;
  • 45) 0.999 999 834 467 343 322 316 8 × 2 = 1 + 0.999 999 668 934 686 644 633 6;
  • 46) 0.999 999 668 934 686 644 633 6 × 2 = 1 + 0.999 999 337 869 373 289 267 2;
  • 47) 0.999 999 337 869 373 289 267 2 × 2 = 1 + 0.999 998 675 738 746 578 534 4;
  • 48) 0.999 998 675 738 746 578 534 4 × 2 = 1 + 0.999 997 351 477 493 157 068 8;
  • 49) 0.999 997 351 477 493 157 068 8 × 2 = 1 + 0.999 994 702 954 986 314 137 6;
  • 50) 0.999 994 702 954 986 314 137 6 × 2 = 1 + 0.999 989 405 909 972 628 275 2;
  • 51) 0.999 989 405 909 972 628 275 2 × 2 = 1 + 0.999 978 811 819 945 256 550 4;
  • 52) 0.999 978 811 819 945 256 550 4 × 2 = 1 + 0.999 957 623 639 890 513 100 8;
  • 53) 0.999 957 623 639 890 513 100 8 × 2 = 1 + 0.999 915 247 279 781 026 201 6;
  • 54) 0.999 915 247 279 781 026 201 6 × 2 = 1 + 0.999 830 494 559 562 052 403 2;

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:


0.000 000 000 742 147 676 637 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 637 3(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:


0.000 000 000 742 147 676 637 3(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 -0.000 000 000 742 147 676 637 3 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