-0.000 000 000 742 147 676 658 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 658 2(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
-0.000 000 000 742 147 676 658 2(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:

|-0.000 000 000 742 147 676 658 2| = 0.000 000 000 742 147 676 658 2


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: 0.000 000 000 742 147 676 658 2.

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) 0.000 000 000 742 147 676 658 2 × 2 = 0 + 0.000 000 001 484 295 353 316 4;
  • 2) 0.000 000 001 484 295 353 316 4 × 2 = 0 + 0.000 000 002 968 590 706 632 8;
  • 3) 0.000 000 002 968 590 706 632 8 × 2 = 0 + 0.000 000 005 937 181 413 265 6;
  • 4) 0.000 000 005 937 181 413 265 6 × 2 = 0 + 0.000 000 011 874 362 826 531 2;
  • 5) 0.000 000 011 874 362 826 531 2 × 2 = 0 + 0.000 000 023 748 725 653 062 4;
  • 6) 0.000 000 023 748 725 653 062 4 × 2 = 0 + 0.000 000 047 497 451 306 124 8;
  • 7) 0.000 000 047 497 451 306 124 8 × 2 = 0 + 0.000 000 094 994 902 612 249 6;
  • 8) 0.000 000 094 994 902 612 249 6 × 2 = 0 + 0.000 000 189 989 805 224 499 2;
  • 9) 0.000 000 189 989 805 224 499 2 × 2 = 0 + 0.000 000 379 979 610 448 998 4;
  • 10) 0.000 000 379 979 610 448 998 4 × 2 = 0 + 0.000 000 759 959 220 897 996 8;
  • 11) 0.000 000 759 959 220 897 996 8 × 2 = 0 + 0.000 001 519 918 441 795 993 6;
  • 12) 0.000 001 519 918 441 795 993 6 × 2 = 0 + 0.000 003 039 836 883 591 987 2;
  • 13) 0.000 003 039 836 883 591 987 2 × 2 = 0 + 0.000 006 079 673 767 183 974 4;
  • 14) 0.000 006 079 673 767 183 974 4 × 2 = 0 + 0.000 012 159 347 534 367 948 8;
  • 15) 0.000 012 159 347 534 367 948 8 × 2 = 0 + 0.000 024 318 695 068 735 897 6;
  • 16) 0.000 024 318 695 068 735 897 6 × 2 = 0 + 0.000 048 637 390 137 471 795 2;
  • 17) 0.000 048 637 390 137 471 795 2 × 2 = 0 + 0.000 097 274 780 274 943 590 4;
  • 18) 0.000 097 274 780 274 943 590 4 × 2 = 0 + 0.000 194 549 560 549 887 180 8;
  • 19) 0.000 194 549 560 549 887 180 8 × 2 = 0 + 0.000 389 099 121 099 774 361 6;
  • 20) 0.000 389 099 121 099 774 361 6 × 2 = 0 + 0.000 778 198 242 199 548 723 2;
  • 21) 0.000 778 198 242 199 548 723 2 × 2 = 0 + 0.001 556 396 484 399 097 446 4;
  • 22) 0.001 556 396 484 399 097 446 4 × 2 = 0 + 0.003 112 792 968 798 194 892 8;
  • 23) 0.003 112 792 968 798 194 892 8 × 2 = 0 + 0.006 225 585 937 596 389 785 6;
  • 24) 0.006 225 585 937 596 389 785 6 × 2 = 0 + 0.012 451 171 875 192 779 571 2;
  • 25) 0.012 451 171 875 192 779 571 2 × 2 = 0 + 0.024 902 343 750 385 559 142 4;
  • 26) 0.024 902 343 750 385 559 142 4 × 2 = 0 + 0.049 804 687 500 771 118 284 8;
  • 27) 0.049 804 687 500 771 118 284 8 × 2 = 0 + 0.099 609 375 001 542 236 569 6;
  • 28) 0.099 609 375 001 542 236 569 6 × 2 = 0 + 0.199 218 750 003 084 473 139 2;
  • 29) 0.199 218 750 003 084 473 139 2 × 2 = 0 + 0.398 437 500 006 168 946 278 4;
  • 30) 0.398 437 500 006 168 946 278 4 × 2 = 0 + 0.796 875 000 012 337 892 556 8;
  • 31) 0.796 875 000 012 337 892 556 8 × 2 = 1 + 0.593 750 000 024 675 785 113 6;
  • 32) 0.593 750 000 024 675 785 113 6 × 2 = 1 + 0.187 500 000 049 351 570 227 2;
  • 33) 0.187 500 000 049 351 570 227 2 × 2 = 0 + 0.375 000 000 098 703 140 454 4;
  • 34) 0.375 000 000 098 703 140 454 4 × 2 = 0 + 0.750 000 000 197 406 280 908 8;
  • 35) 0.750 000 000 197 406 280 908 8 × 2 = 1 + 0.500 000 000 394 812 561 817 6;
  • 36) 0.500 000 000 394 812 561 817 6 × 2 = 1 + 0.000 000 000 789 625 123 635 2;
  • 37) 0.000 000 000 789 625 123 635 2 × 2 = 0 + 0.000 000 001 579 250 247 270 4;
  • 38) 0.000 000 001 579 250 247 270 4 × 2 = 0 + 0.000 000 003 158 500 494 540 8;
  • 39) 0.000 000 003 158 500 494 540 8 × 2 = 0 + 0.000 000 006 317 000 989 081 6;
  • 40) 0.000 000 006 317 000 989 081 6 × 2 = 0 + 0.000 000 012 634 001 978 163 2;
  • 41) 0.000 000 012 634 001 978 163 2 × 2 = 0 + 0.000 000 025 268 003 956 326 4;
  • 42) 0.000 000 025 268 003 956 326 4 × 2 = 0 + 0.000 000 050 536 007 912 652 8;
  • 43) 0.000 000 050 536 007 912 652 8 × 2 = 0 + 0.000 000 101 072 015 825 305 6;
  • 44) 0.000 000 101 072 015 825 305 6 × 2 = 0 + 0.000 000 202 144 031 650 611 2;
  • 45) 0.000 000 202 144 031 650 611 2 × 2 = 0 + 0.000 000 404 288 063 301 222 4;
  • 46) 0.000 000 404 288 063 301 222 4 × 2 = 0 + 0.000 000 808 576 126 602 444 8;
  • 47) 0.000 000 808 576 126 602 444 8 × 2 = 0 + 0.000 001 617 152 253 204 889 6;
  • 48) 0.000 001 617 152 253 204 889 6 × 2 = 0 + 0.000 003 234 304 506 409 779 2;
  • 49) 0.000 003 234 304 506 409 779 2 × 2 = 0 + 0.000 006 468 609 012 819 558 4;
  • 50) 0.000 006 468 609 012 819 558 4 × 2 = 0 + 0.000 012 937 218 025 639 116 8;
  • 51) 0.000 012 937 218 025 639 116 8 × 2 = 0 + 0.000 025 874 436 051 278 233 6;
  • 52) 0.000 025 874 436 051 278 233 6 × 2 = 0 + 0.000 051 748 872 102 556 467 2;
  • 53) 0.000 051 748 872 102 556 467 2 × 2 = 0 + 0.000 103 497 744 205 112 934 4;
  • 54) 0.000 103 497 744 205 112 934 4 × 2 = 0 + 0.000 206 995 488 410 225 868 8;

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:


0.000 000 000 742 147 676 658 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 658 2(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:


0.000 000 000 742 147 676 658 2(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 -0.000 000 000 742 147 676 658 2 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