-0.000 000 000 659 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 659(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
-0.000 000 000 659(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:

|-0.000 000 000 659| = 0.000 000 000 659


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: 0.000 000 000 659.

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) 0.000 000 000 659 × 2 = 0 + 0.000 000 001 318;
  • 2) 0.000 000 001 318 × 2 = 0 + 0.000 000 002 636;
  • 3) 0.000 000 002 636 × 2 = 0 + 0.000 000 005 272;
  • 4) 0.000 000 005 272 × 2 = 0 + 0.000 000 010 544;
  • 5) 0.000 000 010 544 × 2 = 0 + 0.000 000 021 088;
  • 6) 0.000 000 021 088 × 2 = 0 + 0.000 000 042 176;
  • 7) 0.000 000 042 176 × 2 = 0 + 0.000 000 084 352;
  • 8) 0.000 000 084 352 × 2 = 0 + 0.000 000 168 704;
  • 9) 0.000 000 168 704 × 2 = 0 + 0.000 000 337 408;
  • 10) 0.000 000 337 408 × 2 = 0 + 0.000 000 674 816;
  • 11) 0.000 000 674 816 × 2 = 0 + 0.000 001 349 632;
  • 12) 0.000 001 349 632 × 2 = 0 + 0.000 002 699 264;
  • 13) 0.000 002 699 264 × 2 = 0 + 0.000 005 398 528;
  • 14) 0.000 005 398 528 × 2 = 0 + 0.000 010 797 056;
  • 15) 0.000 010 797 056 × 2 = 0 + 0.000 021 594 112;
  • 16) 0.000 021 594 112 × 2 = 0 + 0.000 043 188 224;
  • 17) 0.000 043 188 224 × 2 = 0 + 0.000 086 376 448;
  • 18) 0.000 086 376 448 × 2 = 0 + 0.000 172 752 896;
  • 19) 0.000 172 752 896 × 2 = 0 + 0.000 345 505 792;
  • 20) 0.000 345 505 792 × 2 = 0 + 0.000 691 011 584;
  • 21) 0.000 691 011 584 × 2 = 0 + 0.001 382 023 168;
  • 22) 0.001 382 023 168 × 2 = 0 + 0.002 764 046 336;
  • 23) 0.002 764 046 336 × 2 = 0 + 0.005 528 092 672;
  • 24) 0.005 528 092 672 × 2 = 0 + 0.011 056 185 344;
  • 25) 0.011 056 185 344 × 2 = 0 + 0.022 112 370 688;
  • 26) 0.022 112 370 688 × 2 = 0 + 0.044 224 741 376;
  • 27) 0.044 224 741 376 × 2 = 0 + 0.088 449 482 752;
  • 28) 0.088 449 482 752 × 2 = 0 + 0.176 898 965 504;
  • 29) 0.176 898 965 504 × 2 = 0 + 0.353 797 931 008;
  • 30) 0.353 797 931 008 × 2 = 0 + 0.707 595 862 016;
  • 31) 0.707 595 862 016 × 2 = 1 + 0.415 191 724 032;
  • 32) 0.415 191 724 032 × 2 = 0 + 0.830 383 448 064;
  • 33) 0.830 383 448 064 × 2 = 1 + 0.660 766 896 128;
  • 34) 0.660 766 896 128 × 2 = 1 + 0.321 533 792 256;
  • 35) 0.321 533 792 256 × 2 = 0 + 0.643 067 584 512;
  • 36) 0.643 067 584 512 × 2 = 1 + 0.286 135 169 024;
  • 37) 0.286 135 169 024 × 2 = 0 + 0.572 270 338 048;
  • 38) 0.572 270 338 048 × 2 = 1 + 0.144 540 676 096;
  • 39) 0.144 540 676 096 × 2 = 0 + 0.289 081 352 192;
  • 40) 0.289 081 352 192 × 2 = 0 + 0.578 162 704 384;
  • 41) 0.578 162 704 384 × 2 = 1 + 0.156 325 408 768;
  • 42) 0.156 325 408 768 × 2 = 0 + 0.312 650 817 536;
  • 43) 0.312 650 817 536 × 2 = 0 + 0.625 301 635 072;
  • 44) 0.625 301 635 072 × 2 = 1 + 0.250 603 270 144;
  • 45) 0.250 603 270 144 × 2 = 0 + 0.501 206 540 288;
  • 46) 0.501 206 540 288 × 2 = 1 + 0.002 413 080 576;
  • 47) 0.002 413 080 576 × 2 = 0 + 0.004 826 161 152;
  • 48) 0.004 826 161 152 × 2 = 0 + 0.009 652 322 304;
  • 49) 0.009 652 322 304 × 2 = 0 + 0.019 304 644 608;
  • 50) 0.019 304 644 608 × 2 = 0 + 0.038 609 289 216;
  • 51) 0.038 609 289 216 × 2 = 0 + 0.077 218 578 432;
  • 52) 0.077 218 578 432 × 2 = 0 + 0.154 437 156 864;
  • 53) 0.154 437 156 864 × 2 = 0 + 0.308 874 313 728;
  • 54) 0.308 874 313 728 × 2 = 0 + 0.617 748 627 456;

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:


0.000 000 000 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0100 1001 0100 0000 00(2)

6. Positive number before normalization:

0.000 000 000 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0100 1001 0100 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:


0.000 000 000 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0100 1001 0100 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0100 1001 0100 0000 00(2) × 20 =

1.0110 1010 0100 1010 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):
1.0110 1010 0100 1010 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. 011 0101 0010 0101 0000 0000 =

011 0101 0010 0101 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) =
011 0101 0010 0101 0000 0000


Decimal number -0.000 000 000 659 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 0101 0010 0101 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